Example 1

Table 1 shows the sales revenue and advertisement expenses of a company for the past 10 months. Find the coefficient of correlation between sales and advertisement.

Table 1

Sales and advertisement for 10 months.

Month Jan Feb Mar Apr May Jun Jul Aug Sept OctAdvertisement (in Thousand Rupees)

10 11 12 13 11 10 9 10 11 14

Sales (in thousand rupees)

110 120 115 128 137 145 150 130 120 115

Solution : = −960

1850.2434

=-0.51

Example 2

Table 2 exhibits consumption of electricity (in gigawatt hours) by industry, agriculture, domestic and commercial sectors in India. Compute the range of electricity consumption for these sectors.

Table 2

Consumption of electricity (in gigawatt hours) by industry, agriculture, domestic and commercial sectors in India from 1994-1995 to 2004-2005.

Year Industry Agriculture Domestic Commercial1994-19951995-19961996-19971997-19981998-19991999-20002000-20012001-20022002-20032003-20042004-2005

100,126104,693104,165104,926105,080106,728107,622107,296114,959124,573137,589

79,30185,73284,01991,24297,19590,93484,72981,67384,48687,08988,555

47,91551,73355,26760,34664,97370,52075,62979,69483,35589,73695,660

15,97316,99617,51919,36719,79921,1612254524,13925,43728,20131,381

Example 3

In a grocery store, the mean expenditure per customer is Rs. 2000 with a standard deviation of Rs. 300. If a random sample of 50 customers selected, what is the probability that the sample average expenditure per customer is more than Rs. 2080?

Solution : = 1.88

Example 4

For example 3, determine the probability that the sample average expenditure per customer is between Rs. 2040 and Rs. 2080.

Solution : = 0.04699-0.3264

= 0.1435

Example 5

The bottled water segment in India has witness’s rapid growth. Institutional users are responsible for 30% sales in the market. If 100 customers are randomly selected, what is the probability that 25 or more customers are institutional users?

Solution : = 0.3621+0.05000

= 0.8621

Example 6

Unisys.com is one of the most frequented b2b websites. According to WSJ, business partners accessing Unisys spend an average of 65.7 minutes, possibly the longest average time per visit of any b2b website. Assume that the length of a visit on the Unisys website is distributed as a normal random variable with a standard deviation of 15 minutes.

a. What is the probability that a randomly selected visit will last more than 90 minutes?b. What is the probability that a randomly selected visit to Unisys.com will last between 60 and 90 minutes? c. Only 20% of the visits will last less than how many minutes?d. Between what two values (in minutes) symmetrically distributed around the population mean will 90% of

the visits last?

Example 7

By the year 2014-2015, the telephone instrument industry is estimated to grow by 106.20 millions units as compared to 1993-1994 when the total market size was only 3 million units. Bharti Terletech, BPL Telecom, ITI (Indian Telephone Industries, Bharti Systel, Tata Telecom and Godrej Telecom are some of the major players in the market. Bharti Teletech has a market share of 24%. If 200 purchases of telephone instruments are randomly selected, what is the probability that 55 or more are Bharti Teletech customers?

Solution : = 1.16

The z value is obtained is 1.16 and corresponding probability from the normal table is 0.3770. This is the area between z=0 and z=1.16. so, total area less than 1.16 is equal to 0.5000+0.3770=0.8770. Hence, when 200 purchasers of telephone instruments are randomly selected, probability that 55 or more are Bharti Teletech customers is equal to 1-0.8770=0.1230

Example 8

A researcher has taken a random sample of size 70 from a population with a sample mean of 35 and a population standard deviation of 4.62. Construct a 90% confidence interval to estimate the population mean.

Solution : = p(34.091 ≤ µ ≤ 35.909) = 0.90

Example 9

A researcher wants to measure the income level of employees working in a company. The total employee strength of the company is 1200. A random sample of 50 employees reveals that the average income of sampled employees is Rs 15,000. Historical data reveals that the standard deviation of the income of the employee is approximately Rs. 1500. Construct a 99% confidence interval for obtaining the average income of all the employees working in this company.

Solution : = 15,000-534.96 ≤ µ ≤ 15,000+534.96

= 14,465.04 ≤ µ ≤ 15,534.96

Example 10

The personnel department of an organization wants to apply cost-cutting measures for improving efficiency. As the first step, the personnel department wants to curtail telephone expenses incurred by employees. For this, personnel department has taken a random sample of 10 employees and gathered the following data about telephone expenses (in thousand rupees) in the previous year:

10,12,24,23,11,14,15,34,16,23

Construct a 95% confidence interval to estimate the average telephone expenses of the employees in the population.

Solution : = 18.2-5.42 ≤ µ ≤ 18.2+5.42 = 12.78 ≤ µ ≤ 23.62

Example 11

A research company conducted a survey on 300 randomly selected tax payers. It found that out of 300 tax payers, 180 tax payers have filled the “SARAL” form correctly. Construct a 95% confidence interval to estimate the percentage of tax payers who have filled the form correctly in the population.

Solution : = 0.54 ≤ µ ≤ 0.65

Example 12

A population has a standard deviation of 4.2 and sampling error of 2.4. Determine the sample size to estimate the mean of the population with 95% confidence level?

Solution : = n=3.8416 x17.64

5.76 = 67.765.76

= 11.76 = 12 Apprpximately

Example 13

A consumer electronics company wants to determine the job satisfaction levels of its employees. For this, they ask a simple question, are you satisfied with your job? It was estimated before the study that no more than 30% of the employees would answer yes. What should be the sample size for this company to estimate population proportion to ensure 95% confidence level in result, and to be within 0.04 of the true population proportion?

Solution : = 0.80670.0016

= 504.18

Example 14

A multinational company sells through small retail shops. The company has taken a random sample of 75 retail shops. The average sales per day from the sampled shops are computed as Rs. 5000. The population standard deviation is estimated as Rs. 1000. Construct a 90% confidence interval to estimate the mean for the population.

Solution : = 4810.0686 ≤ µ ≤ 5189.9313

Example 15

A mineral company has launched a new 10 litres bottle in the urban market. After sometimes, the firm receives some complaints that the bottles do not contain exactly 10 litres. For verifying this complaint, the company’s investigating team has taken a random sample of 120 bottles from different places. The sample mean is computed as 9.5 and the sample standard deviation is computed as 1.1. Construct a 95% confidence interval to estimate the mean for the population.

Solution : = 9.30318 ≤ µ ≤ 9.69681

Example 16

A distemper manufacturing company has launched a 2 kg bag in the market and decided to price this bag at Rs. 100. The price of this bag varies from town to town. The company researches have taken a random sample from 40 shops located in different towns. The price collected from 40 shops is given in Table . Construct a 99% confidence interval to estimate the mean for the population.

1 1012 1023 1014 1005 1036 1007 998 989 10310 10211 10212 10313 9914 101

15 10216 9917 9818 10119 9720 10221 10122 10223 99

Shop no. Price

Shop No. Price

24 10325 10426 9827 9728 9929 10130 10031 9932 9833 10234 10135 9936 10237 10238 10339 10140 100

Solution : = 99.852 ≤ µ ≤ 101.348

Example 17

A company receives copper plates from a vendor to use as an important part of its machinery. The company had specified that the diameter of the copper plates must be 20 millimeters. Production department of the company has observed that a few of the supplied plates do not meet the specifications. For verifying this, the company researches have taken a random sample of 20 plates. Assuming that the diameters of these plates are normally distributed, construct a 95% confidence interval for estimating the population mean diameter. Diameter of 20 randomly sampled plates is given in table .

Diameter of the copper plates supplied by vendor

19.92 19.90 20.00 20.12 20.11 20.00 19.98 19.99 20.01 20.03

20.06 20.08 19.94 19.96 19.97 20.03 20.05 20.07 20.00 19.99

Solution : = 19.9828 ≤ µ ≤ 20.0382

Example 18

The quality control department of an electric bulb manufacturing company wants to estimate the average life of the electric bulbs. For this, a quality control inspector has taken a random sample of 20 bulbs. Life of these bulbs in hours is given in Table . Assuming that the life of the bulb is normally distributed, construct a 90% confidence interval for estimating the population mean.

Life of 20 randomly sampled bulbs in hours

99 101 98 95 89 97 96 101 98 102 98 99 101 99 96 97 95 103 101 96

Solution : = 96.81 ≤ µ ≤ 99.28

Example 19

The organized sector has a 70% market share in the storage batteries segment in India. Suppose a researcher wants to check this market share. For this purpose, the researcher has taken a random sample of 120 customers of storage batteries. Out of 120 customers, 86 customers have purchased from the organized market. Assuming customers to be normally distributed, construct a 95% confidence interval for estimating the population proportion.

Solution : = 0.6360 ≤ µ ≤ 0.7972

Example 20

The footwear market in India can be divided into three categories: leather products, rubber/PVC products, and canvas products. Market share for leather products is 10%. Suppose a footwear company wants to launch a new expensive product. The company management has decided that it will launch this product only when there is an increase in the market shares of leather products. Company researchers have taken a random sample of 200 customers and found that 28 customers have purchased leather products. Assuming customers to be normally distributed, construct a 90% confidence interval for estimating population proportion.

Solution : = 0.0996 ≤ µ ≤ 0.1803