Here are the solutions:


TCO 6) Horse race time is found to be normally distributed with a mean value of 10 minutes and a standard deviation of 2 minutes. Horses whose race time is in the top 5% will not be eligible to participate in a second round. What is the lowers race time that makes a horse losses his eligibility to participate in a second round? (Points : 6)
13.29
12
6.71
15.29

Answer: 13.29


A Mall manager claims that in average every customer spends $52 per a single visit to the mall. To test this claim, you took a sample of 36 customers and found the sample mean to be $55 and the sample standard deviation to be $12. At alpha = 0.05, test the Mall’s manager claim. Perform an appropriate hypothesis test, showing the necessary calculations and/or explaining the process used to obtain the results. (Points : 20)

Solution:

Null Hypothesis: H0: µ = 52
Alternate Hypothesis H1: µ ≠ 52
Test statistic: z = (55-52)/(12/sqrt(36)) = 1.5
Critical z-values = -1.96, 1.96
Since calculated test statistics does not fall in rejection region so we fail to Reject Null hypothesis. Thus there is enough evidence to support manager’s claim that in average every customer spends $52 per a single visit to the mall.


A bank manager wanted to estimate the mean number of transactions businesses make per month. For a sample of 60 businesses, he found the mean number of transaction per month to be 38 and the standard deviation to be 8.5 transactions.
(a) Find a 95% confidence interval for the mean number of business transactions per month. Show your calculations and/or explain the process used to obtain the interval.

Solution:

95% confidence interval will be
= (38- 1.96*8.5/sqrt(60), 38+ 1.96*8.5/sqrt(60))
= (35.85, 40.15)

(b) Interpret this confidence interval and write a sentence that explains it. (Points : 20)

Solution:

It means that we are 95% confident that the true mean number of transactions per month is between 35.85 and 40.15.



A company’s CEO wanted to estimate the percentage of defective product per shipment. In a sample containing 400 products, he found 25 defective products.
(a) Find a 95% confidence interval for the true proportion of defective product. Show your calculations and/or explain the process used to obtain the interval.

Solution:

p-hat = 25/400 = 0.0625

95% confidence interval for the true proportion of defective product will be

=(0.0625 – 1.96*sqrt(0.0625*(1-0.0625)/400), 0.0625 + 1.96*sqrt(0.0625*(1-0.0625)/400))

= (0.0388, 0.0862)

(b) Interpret this confidence interval and write a sentence that explains it. (Points : 20)

Solution:

It means that we are 95% confident that the true proportion of defective product lies between 0.0388 and 0.0862.


The ages of 10 students are listed in years:{ 18,20,27,24,29,21,28,19,22,28}
(a) Find the mean, median, mode, sample variance, and range.

Mean = 23.6
Median = 23
Mode = 28
Range = 11
Sample variance = 17.16

(b) Do you think that this sample might have come from a normal population? Why or why not? (Points : 20)

Since the mean and the median are approximately the same, this sample might have come from a normal population.

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