# Normal Distributions and Confidence Intervals

1. Five machines produce electronic components. The number of components produced per hour is normally distributed with a mean of 25 and a standard deviation of 4.

a. What percentage of the time does a machine produce more than 27 components per hour?

b. What percentage of the time is the average rate of output of the five machines more than 27 components per hour?

2. In analyzing the operating cost for a huge fleet of delivery trucks, a manager takes a sample of 25 cars and calculates the sample mean and sample standard deviation. Then he finds a 95% confidence interval for the mean cost to be $253 to $320. He reasons that this interval contains the mean operating cost for the fleet of delivery trucks because the sample mean is contained in the interval. Do you agree? How would you interpret this confidence interval?

3. For the Canadian government and casino operators, gambling has provided the basis for a very profitable marriage. Some 14 million visitors come to Canada annually, many to gamble at the casinos. The lion's share of that action is in Central Canada, which has five glitzy casinos—Niagara Falls, Windsor, and Casino Rama near Orillia in Ontario and Hull and Montreal in Quebec. A survey at one of these casinos revealed that Canadian households that gamble spend between $100 and $3,200 annually in gambling. What sample size would be necessary to estimate the average amount spent by Canadian households that visit the casinos so that the estimate of the average is within $150 with 95% confidence?

4. After taking a random sample of 40 customers and asking them the amount of time they spend shopping at a particular store, a 95% confidence interval is computed to be 10 to 32 for the average time spent shopping. An analyst can correctly make which of the following statements?

a. I am 95% confident that the average time that a customer spends shopping is between 10 and 32 minutes.

b. The probability is .95 that the midpoint of the interval is equal to the population mean.

c. The mean of the population is between 10 and 32.

d. If I repeatedly obtained samples of size 40, then 95% the resolving confidence intervals would contain the mean time that customers spend shopping.

5. In August 2000, new car prices were on average 11% lower than they were the year before. In August 2001, the price of new cars was again lower, but not by as much. The prices were approximately 5.6% lower then in 2001. Suppose that an automobile analyst wished to obtain information on how much lower a new medium-sized car was selling for in Richmond, Virginia. The following data are the results of a random sample of 20 new cars that were sold in August 2001. These figures represent the decrease in the 2001-ceiling price (in percentage) compared to the 2000-ceiling price.

4.1 5.3 5.8 6.7 4.8 7.3 6.5 5.1 7.2 5.2

6.7 8.1 3.5 5.2 6.1 4.2 3.2 6.8 3.8 7.1

a. Compare a 90% confidence interval for the mean decrease in the price of a new medium sized car selling in Richmond, Virginia. Interpret the confidence interval.

b. What could you do to obtain a narrower confidence interval?

c. What sample size is necessary to estimate the sample mean with 90% confidence such that it is within .2 of the actual decrease in selling price on a year-to-year basis?

https://brainmass.com/statistics/normal-distribution/normal-distributions-confidence-intervals-530387

#### Solution Preview

Please refer to the attachment for the solution.

1. Five machines produce electronic components. The number of components produced per hour is normally distributed with a mean of 25 and a standard deviation of 4.

a. What percentage of the time does a machine produce more than 27 components per hour?

b. What percentage of the time is the average rate of output of the five machines more than 27 components per hour?

Solution:

a.

z−score=x−μσ=27−254=0.5

Px>27=PZ>0.5=0.5−0.1915=0.3085

The percentage of the time does a machine produce more than 27 components per hour is 30.85%.

b. According to Central Limit Theorem, the mean and the standard deviation for the given sampling distribution can be calculated as follows:

Sample Size, n=5

Sample Mean,μ=μ=25

Sample Standard Deviation,σ=σn=45=1.789

z−score=x−μσ=27−251.789=1.118

Px>27=PZ>1.118=0.5−0.3686=0.1314

The percentage of the time is the average rate of output of the five machines more than 27 components per hour is 13.14%.

2. In analyzing the operating cost for a huge fleet of delivery trucks, a manager takes a sample of 25 cars and calculates the sample mean and sample standard deviation. Then he finds a 95% confidence interval for the mean cost to be $253 to $320. He reasons that this interval contains the mean operating cost for the fleet of delivery trucks because the sample mean is contained ...

#### Solution Summary

The solution determines the normal distributions and confidence intervals.