Question

1.   A researcher conducted a survey on a university campus for a sample of 64 seniors. The research reported that seniors read an average of 3.12 books in the prior academic semester, with a standard deviation of 2.15 books. Determine the probability that the sample mean is:
   a) above 3.45
   b) between 3.38 and 3.58
   c) below 2.94
2.   Al Gore, the Democratic candidate for President of the USA in the 2000 election, believes that the proportion of voters who will vote for a Democrat candidate in the year 2004 presidential elections is 0.65. A sample of 500 voters is selected at random.
a.   Assume that Gore is correct and p = 0.65. What is the sampling distribution of the sample proportion ? Explain.
b.   Find the expected value and the standard deviation of the sample proportion .
c.   What is the probability that the number of voters in the sample who will vote for a Democrat presidential candidate in 2004 will be between 340 and 350?
3.   A random sample of 10 waitresses in Iowa City, Iowa revealed the following hourly earnings (including tips):

$19   18   15   16   18   17   16   18   20   14

If the hourly earnings are normally distributed with a standard deviation of $4.5, estimate with 95% confidence the mean hourly earnings for all waitresses in Iowa City.
4.   The number of cars sold annually by used car salespeople is normally distributed with a standard deviation of 15. A random sample of 400 salespersons was taken and the mean number of cars sold annually was found to be 75. Find the 95% confidence interval estimate of the population mean. Interpret the interval estimate.
5.   Suppose that in a large city, the annual income of real estate agents is normally distributed with a standard deviation of $7,500. A random sample of 16 real estate agents reveals that the mean annual income is $52,000. Determine the 99% confidence interval estimate of mean annual income of all real estate agents in the city.
6.   A medical statistician wants to estimate the average weight loss of people who are on a new diet plan. In a preliminary study, he guesses that the standard deviation of the population of weight losses is about 10 pounds. How large a sample should he take to estimate the mean weight loss to within 2 pounds, with 90% confidence?

Answer 1

1)

Solution :

(Given that ,

mean = = 3.12

standard deviation = = 2.15

n = 64

= 3.12 and

= / n = 2.15 / 64 = 2.15 / 8 = 0.26875

(a)

P( > 3.45) = 1 - P( < 3.45)

= 1 - P(( - ) / < (3.45 - 3.12) / 0.26875)

= 1 - P(z < 1.228)

= 1 - 0.8903

= 0.1097

Probability = 0.1097

(b)

P(3.38 < < 3.58) = P((3.38 - 3.12) / 0.26875 <( - ) / < (3.58 - 3.12) / 0.26875))

= P(0.9674 < Z < 1.7116)

= P(Z < 1.7116) - P(Z < 0.9674)

= 0.9565 - 0.8333

= 0.1232

Probability = 0.1232

(c)

P( < 2.94) = P(( - ) / < (2.94 - 3.12) / 0.26875)

= P(z < -0.67)

= 0.2514

Probability = 0.2514