Question

In: Statistics and Probability

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed...

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 4.1 minutes and the standard deviation was 0.40 minutes.

What is the probability that calls last between 4.1 and 4.8 minutes? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

What is the probability that calls last more than 4.8 minutes? (RoundYour z value to 2 decimal places and final answer to 4 decimal places.)

What is the probability that calls last between 4.8 and 5.5 minutes? (Round Your z value to 2 decimal places and final answer to 4 decimal places.)

What is the probability that calls last between 3.5 and 5.5 minutes? (Round Your z value to 2 decimal places and final answer to 4 decimal places.)

As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 5 percent of the calls. What is this time? (Round Your z value to 2 decimal places and final answer to 2 decimal places.)

Solutions

Expert Solution

Solution:

Given that,

mean =   = 4.1 minutes

standard deviation = = 0.40 minutes

a ) P ( 4.1 < x < 4.8 )

P ( 4.1 - 4.1 / 0.40 ) < ( x -  / ) < ( 4.8 - 4.1 / 0.40 )

P ( 0 / 0.40 < z < 0.7 / 0.40 )

P ( 0 < z < 1.75 )

P (z < 1.75 ) - p ( z < 0 )

Using z table

= 0.9599 - 0.5000

= 0.4599

Probability = 0.4599

b ) P ( x > 4.8 )

= 1 - P ( x < 4.8 )

= 1 - P ( x -  / ) < ( 4.8 - 4.1 / 0.40 )

= 1 - P ( z < 0.7 / 0.40  )

= 1 - P ( z < 1.75 )

Using z table

= 1 - 0.9599

= 0.0401

Probability = 0.0401

c ) P ( 4.8 < x < 5.5 )

P ( 4.8 - 4.1 / 0.40 ) < ( x -  / ) < ( 5.5 - 4.1 / 0.40 )

P ( 0.7 / 0.40  < z < 1.4 / 0.40  )

P ( 1.75 < z < 3.5 )

P (z < 3.5 ) - p ( z < 1.75 )

Using z table

= 0.9998 - 0.9599

= 0.0399

Probability = 0.0399

d ) Using standard normal table,

P(Z > z) = 5%

1 - P(Z < z) = 0.05

P(Z < z) = 1 - 0.05 = 0.95

P(Z < 1.645) = 0.95

z = 1.65

Using z-score formula,

x = z * +

x = 1.65 * 0.40 + 4.1

= 4.76

x = 4.76


Related Solutions

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed...
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 5.1 minutes and the standard deviation was 0.40 minutes. Use Appendix B.3. What is the probability that calls last between 5.1 and 5.8 minutes? (Round your z value to 2 decimal places and final answer to 4 decimal places.) What is the probability that...
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed...
A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 5.1 minutes and the standard deviation was 0.40 minutes. Use Appendix B.3. What is the probability that calls last between 5.1 and 5.8 minutes? (Round your z value to 2 decimal places and final answer to 4 decimal places.) What is the probability that...
A study of long distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut,...
A study of long distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 3.6 minutes and the standard deviation was 0.40 minute. a) What is the probability that calls last between 3.6 and 4.2 minutes? b) What is the porbability that the calls last more than 4.2 minutes? c) What is prob calls last beteen 4.2...
A study of long-distance phone calls made from the corporate offices of the Pepsi Bottling Group...
A study of long-distance phone calls made from the corporate offices of the Pepsi Bottling Group Inc. showed the calls follow the normal distribution. The mean length of time per call was 4.2 minutes and the standard deviation was 0.60 minutes a) What is the probability the calls lasted between 3.5 and 4.1 minutes? b) What is the probability the calls lasted less than 3.4 minutes? c) As part of her report to the president, the director of communications would...
Historically, evening long-distance calls phone calls from a particular city have averages 12 minutes per call....
Historically, evening long-distance calls phone calls from a particular city have averages 12 minutes per call. In a random sample of 20 calls, the sample mean was 10.7 minutes per call with a standard deviation of 4 minutes. Does the sample indicate a change in the mean duration of long distance calls? Test the hypothesis at 10% level of significance, state the hypothesis and report your conclusion. a. estimate the p-value and show your work.
83. Suppose that the length of long distance phone calls, measured in minutes, is known to...
83. Suppose that the length of long distance phone calls, measured in minutes, is known to have an exponential distribution with the average length of a call equal to eight minutes. a. Define the random variable. X= ________________. b. Is X continuous or discrete? c. μ= ________ d. σ= ________ e. Draw a graph of the probability distribution. Label the axes. f. Find the probability that a phone call lasts less than nine minutes. g. Find the probability that a...
A statistical analysis of​ 1,000 long-distance telephone calls made by a company indicates that the length...
A statistical analysis of​ 1,000 long-distance telephone calls made by a company indicates that the length of these calls is normally​ distributed, with a mean of 280280 seconds and a standard deviation of 3030 seconds. Complete parts​ (a) through​ (d). a. What is the probability that a call lasted less than 230230 ​seconds?The probability that a call lasted less than 230230 seconds is . 0478.0478 . ​(Round to four decimal places as​ needed.) b. What is the probability that a...
A statistical analysis of​ 1,000 long-distance telephone calls made by a company indicates that the length...
A statistical analysis of​ 1,000 long-distance telephone calls made by a company indicates that the length of these calls is normally​ distributed, with a mean of 240 seconds and a standard deviation of 30 seconds. Complete parts​ (a) through​ (d). a. What is the probability that a call lasted less than 180 ​seconds? The probability that a call lasted less than 180 seconds is nothing. ​(Round to four decimal places as​ needed.) b. What is the probability that a call...
General This case study is from the Corporate Finance book, chapter 8 in the eBook (pg....
General This case study is from the Corporate Finance book, chapter 8 in the eBook (pg. 653 in the hard copy, page 261 in eBook). The case study will require you to perform a financial analysis and make some capital investment decisions for Bethesda Mining Company. You will need to prepare various operating cash flows in order to perform your analysis and make a recommendation. The case study is also stated below. Please complete the case study using the excel...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT