Stat-I Que Chapterwise with ans

STAT-I que with ans

Chapter:1

State with suitable examples the role played by computer technology in applied statistics and the role of statistics in information technology.

 Asked on 2078 Exam

Solution

Computer technology plays a critical role in applied statistics by enabling the efficient storage, management, analysis, and visualization of large and complex datasets. Some specific examples of the role of computer technology in applied statistics are:

1. Data collection and management: Computer technology allows for the automated collection, storage, and management of large amounts of data. For example, online surveys and data logging systems are commonly used to collect and store data in various fields such as marketing, healthcare, and social sciences.
2. Data analysis and modeling: Statistical software packages such as R, Python, and SAS provide powerful tools for analyzing and modeling complex datasets. These software packages allow for the use of a wide range of statistical techniques, including regression analysis, hypothesis testing, and data visualization.
3. Machine learning and data mining: Machine learning algorithms and data mining techniques are increasingly used to analyze and extract insights from large and complex datasets. These techniques can be used in a variety of fields, including finance, healthcare, and marketing.

On the other hand, statistics plays a critical role in information technology by providing tools and techniques for analyzing and making decisions based on data. Some specific examples of the role of statistics in information technology are:

1. Quality control: Statistical quality control techniques are commonly used in manufacturing and software development to ensure that products meet certain quality standards.
2. Data analysis and decision-making: Statistical techniques are used to analyze and make decisions based on data in a variety of fields, including finance, marketing, and healthcare. For example, statistical models are used to predict customer behavior, identify market trends, and evaluate the effectiveness of medical treatments.
3. Experimental design: Statistical experimental design techniques are used to design experiments and evaluate the impact of various factors on the outcome of the experiment. This is important in fields such as engineering, where experiments are often conducted to test the performance of products and systems.

Define primary data and secondary data and explain the difference between them.

Primary data is the data that is collected for the first time through personal experiences or evidence, particularly for research. It is also described as raw data or first-hand information. The mode of assembling the information is costly, as the analysis is done by an agency or an external organisation, and needs human resources and investment. The investigator supervises and controls the data collection process directly.

Secondary data is a second-hand data that is already collected and recorded by some researchers for their purpose, and not for the current research problem. It is accessible in the form of data collected from different sources such as government publications, censuses, internal records of the organisation, books, journal articles, websites and reports, etc.

Difference Between Primary Data and Seconday Data:

Basis	Primary Data	Secondary Data
Definition	Primary data are those that are collected for the first time.	Secondary data refer to those data that have already been collected by some other person.
Originality	These are original because these are collected by the investigator for the first time.	These are not original because someone else has collected these for his own purpose.
Nature of Data	These are in the form of raw materials.	These are in the finished form.
Reliability and Suitability	These are more reliable and suitable for the enquiry because these are collected for a particular purpose.	These are less reliable and less suitable as someone else has collected the data which may not perfectly match our purpose.
Time and Money	Collecting primary data is quite expensive both in the terms of time and money.	Secondary data requires less time and money; hence it is economical.
Precaution and Editing	No particular precaution or editing is required while using the primary data as these were collected with a definite purpose.	Both precaution and editing are essential as secondary data were collected by someone else for his own purpose.

What do your mean by measurement scale? Describe the different types of measurement scales used in statistics.

The measurement consisting of counting the number of units or parts of units displayed by objects and phenomena is called measurement scale. Following are the different types of measurement scale.

a. Nominal Scale:

It is a smallest and lowest level of measurement scale. It is simply a system of assigning number or the symbol to objects or events to distinguish one from anotehr in order to label them. The symbols or numbers have no numerical meaning. The arthematic operations cannot be used for these numerals.

b) Ordinal Scale:

The second and the lowest level of ordered scale is  the ordinal scale. It is the quantification of items by ranking. In this scale, the numerals are arranged in some order but the gaps between the positions of the numerals are not made equal. It represents quantitative values in ascending or descending order.

c) Interval Scale:

In addition to ordering the data, this scale was equidistant units to measure the difference between the scores. It assumes data has equal intervals. This scale doesn’t have absolute zero but only arbitary zero. Interval scale is the developed from of ordinary scale.

d) Ratio Scale:

Ratio scale is the ideal scale and on extended form of interval scale. It is most powerful scale of measurement. It possesses the characteristics of nominal, ordinal and interval scale. Ratio scale has as absolute zero or ture zero or natural zero of measurement.

Chapter:2

What do you understand by measures of central tendency and dispersion in descriptive statistics? What are the different measures of dispersion? The grouped frequency distribution in the table represents masses of a sample of 50 of the people from the data file, “Brain size”.

Mass(lbs)	1-2	2-3	3-4	4-5	5-6	6-7
Frequency	8	10	15	9	6	2

Compute mean, standard deviation, variance and coefficient of variation and interpret them.

 Asked on 2080 Exam

Measures of Central Tendency:

Measures of central tendency are statistical measures that aim to describe the center or typical value of a dataset. They provide a single value that represents the “average” or central value around which the data points tend to cluster. The three main measures of central tendency are:

1. Mean: The mean, also known as the average, is calculated by summing up all the values in a dataset and then dividing by the number of data points. It is represented by the symbol “μ” for the population mean and “x̄” for the sample mean. The formula for the mean is: Mean (μ or x̄) = (Σx) / n Where Σx represents the sum of all data values, and n is the number of data points.
2. Median: The median is the middle value when the data is sorted in ascending or descending order. If there is an even number of data points, the median is the average of the two middle values. The median is not affected by extreme values (outliers) and is often used for skewed datasets.
3. Mode: The mode is the value that occurs most frequently in a dataset. A dataset can have one mode (unimodal), more than one mode (multimodal), or no mode at all. The mode is useful for identifying the most common value(s) in a dataset, especially for categorical data.

Measures of Dispersion:

Measures of dispersion, also known as measures of variability or spread, provide information about the extent to which data points deviate from the central tendency. They help us understand how scattered or spread out the data is. Some common measures of dispersion include:

1. Range: The range is the simplest measure of dispersion and is calculated as the difference between the maximum and minimum values in a dataset. It provides a rough estimate of the spread of data but is sensitive to outliers.
2. Variance: Variance measures the average of the squared differences between each data point and the mean. It quantifies how data points are distributed around the mean. Population variance (σ²) and sample variance (s²) are used for populations and samples, respectively.

   Population Variance (σ²) = Σ[(x – μ)²] / N Sample Variance (s²) = Σ[(x – x̄)²] / (n – 1)

3. Standard Deviation: The standard deviation is the square root of the variance. It provides a measure of dispersion in the same units as the data. A smaller standard deviation indicates that data points are closer to the mean, while a larger standard deviation indicates greater variability.

   Population Standard Deviation (σ) = √σ² Sample Standard Deviation (s) = √s²

4. Interquartile Range (IQR): The IQR is a measure of dispersion that represents the range between the first quartile (Q1) and the third quartile (Q3) of a dataset after sorting it. It is less sensitive to outliers compared to the range.
5. Mean Absolute Deviation (MAD): MAD is the average of the absolute differences between each data point and the mean. It provides a measure of the average deviation of data from the mean.

Write notes on any two:

1. Nominal and ordinal scale
2. Kurtosis
3. Five number summary

 Asked on 2079 Exam

(i) Nominal and ordinal scale:

Nominal and ordinal scales are two types of categorical scales used in statistics.

Nominal scale: This is the simplest type of scale where data are grouped into categories that do not have any inherent order or ranking. Examples include gender, race, religion, etc. Nominal data can only be described using frequency counts and percentages.

Ordinal scale: This type of scale has categories that are ordered or ranked, but the differences between categories are not equal. Examples include ratings of satisfaction, degree of agreement, etc. Ordinal data can be described using frequency counts and percentages, as well as measures of central tendency like the median.

(ii) Kurtosis:

Kurtosis is a measure of the shape of the distribution of a random variable. It measures the degree to which the distribution is peaked or flat compared to the normal distribution. A distribution with a high kurtosis value has a sharp peak and heavy tails, while a distribution with a low kurtosis value has a flat peak and light tails. There are three types of kurtosis:

• Mesokurtic: This is the normal distribution, where the kurtosis value is 0.
• Leptokurtic: This is a distribution with a high kurtosis value, where the peak is sharp and the tails are heavy.
• Platykurtic: This is a distribution with a low kurtosis value, where the peak is flat and the tails are light.

(iii) Five-number summary:

The five-number summary is a descriptive statistic used to summarize the distribution of a dataset. It consists of five values: minimum, maximum, median, and the first and third quartiles. The first quartile (Q1) is the value below which 25% of the data fall, while the third quartile (Q3) is the value below which 75% of the data fall. The interquartile range (IQR) is the difference between Q3 and Q1, and it gives a measure of the spread of the middle 50% of the data. The five-number summary is particularly useful in identifying outliers and for comparing distributions of different datasets.

The following table shows the marks obtained by 130 students in computer science.

Marks	less than 40	40-50	50-60	60-70	70-80	80 or above
No. of students	14	24	36	25	20	11

1. Find the appropriate measure of central tendency.
2. Compute the minimum marks obtained by the top 20% of students.

 Asked on 2079 Exam

What do you understand by measures of dispersion? What are the different measures of depression? The following table shows the data of time taken (in seconds) for completing each one of the series of 80 similar chemical experiments.

Time(s)	50-60	61-65	66-70	71-75	76-78
Number of experiments	8	12	28	22	10

Calculate mean time experiments, standard deviation, and variance and comment on your result.

 Asked on 2079 Exam

Measures of dispersion are statistical measures that describe the spread or variability of a set of data values. Dispersion is the degree to which data is scattered or spread out around its central tendency (mean, median or mode). Measures of dispersion provide information about the degree of variation in a dataset and can help in comparing different datasets.

Some common measures of dispersion include:

Range: The difference between the maximum and minimum values in a dataset.
Variance: A measure of how far the values in a dataset are from the mean.
Standard deviation: The square root of the variance, and another measure of how far the values in a dataset are from the mean.
Interquartile range: The range between the 25th and 75th percentiles in a dataset, which contains the middle 50% of the data.
Mean absolute deviation: The average distance of each data point from the mean.
These measures of dispersion can be useful in providing a more complete picture of a dataset and can be used in conjunction with measures of central tendency to provide a more comprehensive analysis of the data.

The lifetime of a certain electronic component is a normal random variate with the expectation of 5000 hours and a standard deviation of 100 hours. Compute the probabilities under the following conditions

1. Lifetime of components between 3000 to 6500 hours
2. Lifetime of components between 3000 t0 6500 hours
3. Lifetime of components more than 6000 hours

 Asked on 2079 Exam

We can use the properties of the normal distribution to calculate the probabilities under the given conditions.

Let X be the lifetime of the electronic component, where X ~ N(5000, 100^2).

a) Probability of the lifetime of components between 3000 to 6500 hours:

We need to find P(3000 < X < 6500). We can standardize X as follows:

Z = (X – μ) / σ = (X – 5000) / 100

Now, we need to find P((-2 < Z < 1.5)).

Using a standard normal distribution table or a calculator, we find that P((-2 < Z < 1.5)) is approximately 0.9332.

Therefore, P(3000 < X < 6500) is approximately 0.9332.

b) Probability of the lifetime of components less than 3000 hours:

We need to find P(X < 3000). We can standardize X as follows:

Z = (X – μ) / σ = (X – 5000) / 100

Now, we need to find P(Z < -2).

Using a standard normal distribution table or a calculator, we find that P(Z < -2) is approximately 0.0228.

Therefore, P(X < 3000) is approximately 0.0228.

c) Probability of the lifetime of components more than 6000 hours:

We need to find P(X > 6000). We can standardize X as follows:

Z = (X – μ) / σ = (X – 5000) / 100

Now, we need to find P(Z > 1).

Using a standard normal distribution table or a calculator, we find that P(Z > 1) is approximately 0.1587.

Therefore, P(X > 6000) is approximately 0.1587.

Compute first four moments about arbitrary point 4 from following distribution and describe the characteristics of data

X	2	3	4	5	6
f	1	3	7	2	1

Calculate Q1, D7 and P58 from the following data and interpret the results.

Weight	0-10	10-15	20-25	25-30	30-35	35-40	40-45	45-50	50-60
No, of person	4	8	30	15	13	6	4	4	1

Solution:

Since the classes are unequal so adjustment is to be made before calculating.

Weight	No. of Person (f)	c.f.
0-5	2	2
5-10	2	4
10-15	8	12
15-20	15	27
20-25	15	42
25-30	15	57
30-35	13	70
35-40	6	76
40-45	4	80
45-50	4	84
50-55	0.5	84.5
55-60	0.5	85
	N = 85

Now,

For Q1:

Q1 = ${\left(\frac{N}{4}\right)}^{th}item$

= ${\left(\frac{85}{4}\right)}^{th}item$ = 21.25th item

In c.f column, the value just greater than 25.25 is 27. so, class is 15 – 20

Then,

Q1 = $L+\frac{\frac{N}{4}–c.f.}{f}\times h$

= $15+\frac{21.25–12}{15}\times 5$

= 18.08

For D7:

D7 = ${(7\times \frac{N}{10})}^{th}item$

= ${(7\times \frac{85}{10})}^{th}item$ = 59.5th item

In c.f column, the value just greater than 59.5 is 70. so, class is 30-35

Then,

D7 = $L+\frac{\frac{7N}{10}–c.f.}{f}\times h$

= $30+\frac{59.5–57}{13}\times 5$

= 30.96

For P58:

P58 = ${\left(\frac{58N}{100}\right)}^{th}item$

= ${(7\times \frac{58\times 85}{100})}^{th}item$ = 49.3th item

In c.f column, the value just greater than 49.3 is 57. so, class is 25-30

Then,

P58 = $L+\frac{\frac{58N}{100}–c.f.}{f}\times h$

= $25+\frac{49.3–43}{15}\times 5$

= 26.43

What are different methods of measuring dispersion. Sample of polythene bags from two manufactures, A, B, are tested by a prospective buyer for bursting pressure and the results are as follows.

Bursting Pressure		5-10	10-15	15-20	20-25	25-30	30-35
Number of bags manufactured by	A	2	9	29	54	11	5
	B	9	11	18	32	27	13

Which set of bags has more uniform pressure? If price are the same, Which manufacture’s bags would be preferred by buyer? Use appropriate statistical tool

Range, interquartile range, standard deviation. and variance are the three commonly used measures of dispersion.

Solution:

Let X be the mid value so that $d^{”}=\frac{17.45}{5}$. Let f and f‘ be the number of bags of manufactures A and B respectively.

Class	x	$d^{‘}=\frac{x–17.45}{5}$	f	f‘	fd‘	fd‘2	f‘d‘	f‘d‘2
5-9.9	7.45	-2	2	9	-4	8	-18	36
10-14.9	12.45	-2	9	11	-9	9	-11	11
15-19.9	17.45	0	29	18	0	0	0	0
20-24.9	22.45	1	54	32	54	54	32	32
25-29.9	27.45	2	11	27	22	22	54	108
30-34.9	32.45	3	5	13	15	15	39	117
		Σd‘ = 3	N=10	N‘ = 110	Σfd‘=78	Σfd‘2=160	Σf‘d‘ = 96	Σf‘d‘2=304

For manufacture A:

$Mean\left({\overline{X}}_A\right)=A+\frac{\sum f{d}^{‘}}{N}\times h$

= $17.45+\frac{78}{100}\times 5$

= 21

Standard Deviation(σA) = $h\sqrt{\frac{\sum f{d}^{‘2}}{N}–{\left(\frac{\sum f{d}^{‘}}{N}\right)}^2}$

= $5\times \sqrt{\frac{160}{110}–{\left(\frac{78}{110}\right)}^2}$

= 5 x 0.975

= 4.875

Covariance(A) = $\frac{{\sigma }_A}{{\overline{X}}_A}\times 100$

= $\frac{4.875}{21}\times 100$

= 23.2%

For manufacture B:

$Mean\left({\overline{X}}_B\right)=A+\frac{\sum f{d}^{‘}}{N}\times h$

= $17.45+\frac{96}{100}\times 5$

= 21.8

Standard Deviation(σA) = $h\sqrt{\frac{\sum f{d}^{‘2}}{N}–{\left(\frac{\sum f{d}^{‘}}{N}\right)}^2}$

= $5\times \sqrt{\frac{204}{110}–{\left(\frac{96}{110}\right)}^2}$

= 5 x 1.41

= 7.05

Covariance(A) = $\frac{{\sigma }_A}{{\overline{X}}_A}\times 100$

= $\frac{7.05}{21.8}\times 100$

= 32.34%

Since, the cofficient of variation for A is less than that for B, the set of bags manufactured by A has more uniform pressure.Hence, the buyer would perfer to buy bags manufactured by A.

Another Method:

Brusting Pressure	M	A	B	AM	BM
5-10	7.5	2	9	15	67.5
10-15	12.5	9	11	112.5	137.5
15-20	17.5	29	18	507.5	315
20-25	22.5	54	32	1215	720
		ΣA = 94	ΣB = 70	ΣAM = 1850	ΣBM = 1240

Brusting pressure of Polythene A = $\frac{\sum AM}{\sum A}$

= $\frac{1850}{94}$

= 19.68

Brusting pressure of Polythene B = $\frac{\sum BM}{\sum B}$

= $\frac{1240}{70}$

= 17.68

Since, the brusting pressure of polythene A is greater than of polythene B. So, Buyers perfers the polythene A.

Compute percentile coefficient of kurtosis from the following data and interpret the result.

Hourly wages (Rs)	23-27	28-32	33-37	38-42	43-47	48-52
Number of workers	22	16	9	4	3	1

Solution:

x	f	cf
23-27	22	22
28-32	16	38
33-37	9	47
38-42	4	51
43-47	3	54
48-52	1	55
	N = 55

Now,

Calculating P75:

P75 = $75{\left(\frac{N}{100}\right)}^{th}item$ = $75{\left(\frac{55}{100}\right)}^{th}$ = 41.25th

In cf, the value just greater than 41.25 is 4. so, class  = 33 – 37

Now,

P75 = $L+\frac{\frac{75N}{100}–cf}{f}\times i$

= $33+\frac{41.25–38}{9}\times 4$

= 34.45

Calculating P25:

P25 = $25{\left(\frac{N}{100}\right)}^{th}item$ = $25{\left(\frac{55}{100}\right)}^{th}$ = 13.75th

In cf, the value just greater than 13.755 is 22. so, class  = 23 – 27

Now,

P25 = $L+\frac{\frac{25N}{100}–cf}{f}\times i$

= $23+\frac{13.75–0}{22}\times 4$

= 23.625

Calculating P90:

P90 = $90{\left(\frac{N}{100}\right)}^{th}item$ = $90{\left(\frac{55}{100}\right)}^{th}$ = 49.5th

In cf, the value just greater than 49.5 is 51. so, class  = 38 – 42

Now,

P90 = $L+\frac{\frac{90N}{100}–cf}{f}\times i$

= $23+\frac{49.5–47}{4}\times 4$

= 40.5

Calculating P10:

P10 = $10{\left(\frac{N}{100}\right)}^{th}item$ = $10{\left(\frac{55}{100}\right)}^{th}$ = 5.5th

In cf, the value just greater than 5.5 is 21. so, class  = 23-27

Now,

P10 = $L+\frac{\frac{10N}{100}–cf}{f}\times i$

= $23+\frac{5.5–0}{22}\times 4$

= 24

Now,

Percentile cofficient of Kurtosis

= $\frac{P_{75}–P_{25}}{2(P_{90}–P_{10})}$

= $\frac{34.45–23.625}{2(40.5–24)}$

= $\frac{10.825}{33}$

= 0.33

A certain machine makes electrical resistors having mean resistance of 40 ohms and standard deviations of 2 ohms. Assuming that the resistance follows a normal distribution.

1. What percentage of resistors will have a resistance exceeding 43 ohms?
2. What percentage of registers will have a resistance between 30 ohms to 45 ohms?

X = resistance of resistors such that X ~ (μ, σ2)

Given,

μ = 40

σ = 2

σ2 = 4

x ~ (40, 4)

Let z = $\frac{x–\mu }{\sigma }$ be the standard normal variate.

Solution i):

p(x > 43)

= $p(z>\frac{x–\mu }{\sigma })$

= $p(z>\frac{43–40}{2}$

= p(z > 1.5)

= 0.5 – p(0 < z < 1.5)

= 0.5 – 0.4332

= 0.0668

Solution ii):

p(30 ≤ x ≤ 45)

= $p(\frac{30-40}{2}\le z\le \frac{45–40}{2})$

= p(-5 ≤ z ≤ 2.5)

= p(-5 ≤ z ≤ 0) + p(0 ≤ z ≤ 2.5)

= p(0 ≤ z ≤ 5) + p(0 ≤ z ≤ 2.5)

= 0.5 + 0.4938

= 0.9938

Calculate Q3, D6, and P80 from the following data and interpret the results.

Respiratory rate	10	15	20	25	30	35	40	45	50
No. of Person	8	12	36	25	28	18	9	12	6

Solution:

x	f	cf
10	8	8
15	12	20
20	36	56
25	25	81
30	28	109
35	18	127
40	9	136
45	12	148
50	6	154
	N = 154

Now,

Q3 = $3{\left(\frac{N+1}{4}\right)}^{th}$ item

= $3\times \frac{155}{4}$ = 116.25

In c,f column, the value just greater than 116.25 is 127. ∴ Q3 = 35

Also,

D6 = $6{\left(\frac{N+1}{10}\right)}^{th}$ item

= $6\times \frac{155}{10}$ = 93th

In c,f column, the value just greater than 93 is 109. ∴ D6 = 30

Also,

P80 = $80{\left(\frac{N+1}{100}\right)}^{th}$ item

= $80\times \frac{155}{100}$ = 124th

In c,f column, the value just greater than 124 is 127. ∴ P80 = 35

Measurement of computer chip’s thickness (in monometers) is recorded below.

Thickness of chips (in nanometers)	34-39	39-44	44-49	49-54	54-59	Total
Number of computers	3	11	16	25	5	60

Find the mode of thickness of computer chips and interpret the result.

Solution:

Here,

Thickness	No. of computer(f)
34-39	3
39-44	11
44-49	16
49-54	25
54-59	5
	N = 60

Here, 49-54 class has highest frequence

Therefore, modal class = 49-54

Now, mode = $L+\frac{f_1–f_0}{2f_1–f_0–f_2}$

= $49+\frac{25–16}{2\times 25–16–5}$

= 49.31

Distinguish between absolute and relative measure of dispersion. Two computer manufacturers A and B compete for profitable and prestigious contract. In their rivalry, each claim that their computer a consistent. For this it was decided to start execution of the same program simultaneously on 50 computers of each company and recorded the time as given below.

Time (in seconds)		0-2	2-4	4-6	6-8	8-10	10-12
Nature of computers manufactured by	A	5	16	13	7	5	4
	B	2	7	12	19	9	1

Which company’s computer is more consistent?

Measure of dispersion is descriptive statistical measure used to measure the variation or spread or scatteriess in the data set.

There are two types of measures of dispersion i.e absolute and relative measure of dispersion. The difference between them are:

Absolute Measure	Relative Measure
A measure of dispersion is saied to be an absolute if it is expressed interms of original units of data.	A measure of dispersion is saied to be relative if it is independent of original units of data and it is simply a cofficient.
Some absolute measures: Range Quartile deviation Mean deviation Standard deviation	Some relative measures are: Cofficient of range Cofficient of Quartile deviation Cofficient of Mean deviation Cofficient of variation
Absolute measures are in fixed value	Relative measures may be in ratio or percentage

Solution:

For Company A

Time(x)	No of Comp.(f)	m	fm	fm2
0-2	5	1	5	5
2-4	16	3	48	144
4-6	13	5	65	325
6-8	7	7	49	343
8-10	5	9	45	405
10-12	4	11	44	484
	N = 50		Σfm = 256	Σfm2 = 1706

Now,

$\overline{X}=\frac{\sum fm}{N}=\frac{256}{50}=5.12$

Also, Standard deviation(σ) is

$\sigma =\sqrt{\frac{\sum f{m}^2}{N}–(\frac{\sum fm}{N}{)}^2}$

$=\sqrt{\frac{1706}{50}–(5.12{)}^2}$

= 2.812

Also, Covariance (C.V) = $\frac{\sigma }{\overline{X}}\times 100$

$=\frac{2.812}{5.12}\times 100$

= 54.92%

For Company B

Time(x)	No of Comp.(f)	m	fm	fm2
0-2	2	1	2	2
2-4	7	3	21	63
4-6	12	5	60	300
6-8	19	7	133	931
8-10	9	9	81	729
10-12	1	11	11	121
	N = 50		Σfm = 308	Σfm2 = 2146

Now,

$\overline{X}=\frac{\sum fm}{N}=\frac{308}{50}=6.16$

Also, Standard deviation(σ) is

$\sigma =\sqrt{\frac{\sum f{m}^2}{N}–(\frac{\sum fm}{N}{)}^2}$

$=\sqrt{\frac{2146}{50}–(6.16{)}^2}$

= 1.725

Also, Covariance (C.V) = $\frac{\sigma }{\overline{X}}\times 100$

$=\frac{1.725}{6.16}\times 100$

= 28%

Since, the standard deviation and covarience of company B is less than that of A. So, Company B is considered more consistent.

The length of power failure in minute are recorded in the following table.

Power failure time	22	23	24	25	26	27	28	Total
Frequency	2	5	7	10	4	3	2	33

Find Q3, D2 and P40 and interpret the results.

Solution:

x	f	cf
22	2	2
23	5	7
24	7	14
25	10	24
26	4	28
27	3	31
28	2	33
	N = 33

Solution For Q3:

Now, Q3 class  = $3{\left(\frac{N+1}{4}\right)}^{th}item$

= ${\left(\frac{3\times 34}{4}\right)}^{th}$ = 25.5th

In cf column, the value just greater than 25.5

It mean 3/4 or 75% of power failure time has value less than or equal to 26.

Solution For D2:

Now, D2 class  = $2{\left(\frac{N+1}{10}\right)}^{th}item$

= ${\left(\frac{2\times 34}{10}\right)}^{th}$ = 6.8th

In cf, the value just greater than 6.8 is 7. so, D2 = 23

Solution For P40:

Now, P40 class  = $40{\left(\frac{N+1}{100}\right)}^{th}item$

= ${\left(\frac{40\times 34}{100}\right)}^{th}$ = 13.6th

In cf, the value just greater than 13.6 is 14. so, P40 = 24

It means 40% of power failure time has value less than or equal to 24.

The following table gives the installation time (in minutes) for hardware on 50 different computers.

Installation Time	0-10	10-20	20-30	30-40	40-50	Total
Number of computers	4	–	10	–	10	50

If the average installation time is 30.2 minutes, find missing frequencies.

Let x and y be the missing frequencies. Now, Let’s construct the table:

Given, x̄ = 30.2

class(x)	No.(f)	m	fm
0-10	4	5	20
10-20	x	15	15x
20-30	10	25	250
30-40	y	35	35y
40-50	10	45	450
	N = 50 = 24 + x + y		Σfm = 720 + 5x + 35y

Now, from table

24 + x + y = 50

x + y = 26  ——— (i)

Also,

15x  + 35y + 720 = 30.2 x 50

15x + 35y = 790 ———–(ii)

Solving equation i) and ii)

$\frac{\begin{array}{c}15x+15y=390\\ 15x+35y=790\end{array}}{–20y=–400}$

y = 20

When, y = 20

x + 20 = 26

x = 6

The missing frequencies are 20 and 6

What are the roles of measure of dispersion in descriptive statistics? Following table gives the frequency distribution of thickness of computer chips (in nanometer) manufactured by two companies.

Thickness of computer chips		5	10	15	20	25	30
Number of chips	Company A	10	15	24	20	18	13
	company B	12	18	20	22	24	4

Measure of dispression is a descriotive statistical measure used to measure the variation or spreed or scatter less in the data set.

It gives additional information that enables to judge the reliability of measure of central tendency.

The measure of dispersion is important because it determines the margin of error. You’ll have when measuring the central tendency.

Solution Part:

For Company A:

x	f	fx	fx2
5	10	50	250
10	15	150	1500
15	24	360	5400
20	20	400	10000
25	18	450	11250
30	13	390	11700
	N = 100	Σfx = 1800	Σfx2 = 40100

Now,

$\overline{x}=\frac{\sum fx}{N}$ = $\frac{1800}{100}$ = 18

Standard deviation(σ) = $\sqrt{\frac{\sum f{x}^2}{N}–{\left(\frac{\sum fx}{N}\right)}^2}$

= $\sqrt{\frac{40100}{100}–{\left(\frac{1800}{100}\right)}^2}$

= 8.77

And, Co-varience (c.v) = $\frac{\sigma }{\overline{x}}\times 100$

= $\frac{8.77}{18}\times 100$

= 48.72%

For Company b:

x	f	fx	fx2
5	12	60	300
10	18	180	1800
15	20	300	4500
20	22	440	8800
25	24	600	15000
30	4	120	3600
	N = 100	Σfx = 1700	Σfx2 = 34000

Now,

$\overline{x}=\frac{\sum fx}{N}$ = $\frac{1700}{100}$ = 17

Standard deviation(σ) = $\sqrt{\frac{\sum f{x}^2}{N}–{\left(\frac{\sum fx}{N}\right)}^2}$

= $\sqrt{\frac{34000}{100}–{\left(\frac{1700}{100}\right)}^2}$

= 7.14

And, Co-varience (c.v) = $\frac{\sigma }{\overline{x}}\times 100$

= $\frac{7.14}{17}\times 100$

= 42%

Since, the standard deviation and covarience of company B is less than that of company A. so, company B is considered more consistent in terms of thickness of computer chip.

Chapter-3

Define conditional probability. A problem of statistics is given to three students. A, B, and C whose chances of solving the problem are in a ratio of 2:3:5. Find the probability that (i) non of them solve the problem (ii) the problem will be solved.

 Asked on 2079 Exam

Conditional probability is the probability of an event occurring given that another event has occurred. It is denoted by P(A | B), where A and B are events.

In the given problem, let the probability of solving the problem by students A, B, and C be P(A), P(B), and P(C) respectively. It is given that their chances of solving the problem are in a ratio of 2:3:5, which means:

P(A) : P(B) : P(C) = 2 : 3 : 5

We can find the probability of each student not solving the problem by subtracting their probabilities of solving the problem from 1, as follows:

P(not A) = 1 – P(A) = 1 – (2 / 10) = 0.8 P(not B) = 1 – P(B) = 1 – (3 / 10) = 0.7 P(not C) = 1 – P(C) = 1 – (5 / 10) = 0.5

(i) To find the probability that none of them solve the problem:

we can use the multiplication rule of probability, which states that the probability of two or more independent events occurring together is the product of their individual probabilities. Since the probabilities of the three students solving the problem are independent of each other, we can find the probability that none of them solve the problem by multiplying their probabilities of not solving the problem, as follows:

P(none of them solve the problem) = P(not A) * P(not B) * P(not C) = 0.8 * 0.7 * 0.5 = 0.28

Therefore, the probability that none of them solve the problem is 0.28.

(ii) To find the probability that the problem will be solved:

we can use the complement rule of probability, which states that the probability of an event occurring is equal to 1 minus the probability of the event not occurring. In this case, the event of the problem being solved is the complement of the event of none of them solving the problem. Therefore, we can find the probability that the problem will be solved as follows:

P(the problem will be solved) = 1 – P(none of them solve the problem) = 1 – 0.28 = 0.72

Therefore, the probability that the problem will be solved is 0.72.

Two batsmen A and B made the following runs in a series of cricket matches.

A	10	0	56	80	24
B	36	37	45	28	29

Who is a more consistent player, and why?

 Asked on 2079 Exam

If 50 image of your website, 10 have black and white image, and their average scanned image occupies with 2.5 megabytes of memory. The total image occupies by the entire work 281 megabytes. Find the average occupies megabytes of those color images.

Given,

Total size/size of all images (Wa) = 281 megabytes

Total images (N) = 50

Size of black and white images (WBW) = 2.5 megabytes

Number of black and white images (New) = 10

Now,

Total size of black and white images (Wt) = Wbw x Nbw = 2.5 x 10 = 25 megabytes

For color images,

Total number of color images (Nc) = N – New = 50 – 10 = 40

Size of color images (Wc) = Wa – Wt = 281 – 25 = 256 megabytes

Average store occupation (Xw) = Wc/Nc = 256/40 = 6.4 megabytes

Hence, the average storage occupancy by color images is 6.4 megabytes.

Define a random variable. For the following bi-variants probability distribution of X and Y , find

1. marginal probability mass function of X and Y ,
2. P(x≤1, Y=2),
3. P(X≤1)

X/Y	1	2	3	4	5	6
0	0	0	1/32	2/32	2/32	3/32
1	1/16	1/16	1/8	1/8	1/8	1/8
2	1/32	1/32	1/64	1/64	1/64	1/64

A variable whose value id determined by the outcome of a random experiment is called a random variable.

Given,

xy	1	2	3	4	5	6	P(X=x)
0	0	0	1/32	2/32	2/32	3/32	8/32
1	1/16	1/16	1/8	1/8	1/8	1/8	5/8
2	1/32	1/32	1/64	1/64	1/64	1/64	1/8
P(Y=y)	3/32	3/32	1/64	13/64	13/64	13/64	sum = 1

Solution i):

Marginal pmf of x = p(X = xi) = ${\sum }_{j=1}^{n}p\left(x_i{y}_i\right)$

∴P(X = 0) = P(X=0, Y=1) + P(X = 0, Y = 2) + P(X = 0, Y = 3) + P(X = 0, Y = 4) + P(X = 0, Y = 5) + P(X = 0, Y = 6)

= 8/32

∴P(X = 1) = P(X=1, Y=1) + P(X = 1, Y = 2) + P(X = 1, Y = 3) + P(X = 1, Y = 4) + P(X = 1, Y = 5) + P(X = 1, Y = 6)

= 5/8

∴P(X = 2) = P(X=2, Y=1) + P(X = 2, Y = 2) + P(X = 2, Y = 3) + P(X = 2, Y = 4) + P(X = 2, Y = 5) + P(X = 2, Y = 6)

= 1/8

Also, Marginal pmf of Y = P(Y=yi) = ${\sum }_{i=1}^{n}p\left(x_i{y}_i\right)$

∴P(Y = 1) = P(X=0, Y=1) + P(X = 1, Y = 1) + P(X = 1, Y = 2)

= 3/32

Similarly,

P(Y = 2) = 3/32

P(Y = 3) = 11/64

P(Y = 4) = 13/64

P(Y = 5) = 13/64

P(Y = 6) = 15/64

Solution ii):

P(X ≤ 1, Y = 2)

= P(X = 0, Y= 2) + P(X = 1, Y = 2)

= 0 + 1/16

= 1/16

Solution ii)

P(X ≤ 1)

= P(X = 0) + P(X = 1)

= 8/32 + 5/8

= 7/8

Message arrives at an electronic message center at random times, with an average of 9 messages per hour.

1. What is the probability of receiving at least four messages during the next hour?
2. What is the probability of receiving at most three messages during the next hour?

let, x = no of messages arriving per hour

Given, Average no. of messages per hour (λ) = 9

we know, p(x) = $\frac{e^{-\lambda }.{\lambda }^x}{x!}$

Solution a):

Probability of receiving at least four messages during the next how,

p(x ≥ 4) = 1 – p(x ≤ 4)

= 1 – {P(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)}

= $1–\{\frac{e^{-9}{.9}^0}{0!}+\frac{e^{-9}{.9}^1}{1!}+\frac{e^{-9}{.9}^2}{2!}+\frac{e^{-9}{.9}^3}{3!}\}$

= 1 – {0.00123 + 0.0011 + 0.0149 + 0.00499}

= 0.9778

Solution b):

Probability of receiving at least messages during the next hour

p(x ≤ 3) = p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)

= 0.00123 + 0.0011 + 0.0149 + 0.00499

= 0.0222