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Confidence interval and Normal distributions

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In a class there are 60 students. The mean exam score is 70 and the
standard deviation is 5. (Be Precise in your answers!) Assume symmetric
distribution of the scores.

(i) What can you say about the number of students that have scores between
55 and 85?

(ii) What can you say about the number of students who have scores between
60 and 80?

4) Let Z have a standard normal distribution. Find the following
probabilities:
a) P(Z is between -2.52 and 2.67)

b) P( Z is more than -1.26)

5) A gas company president for a particular city is interested in the
proportion of homes heated by gas. Historically, the proportion of homes
heated by gas has been 65%. A sample of 75 home was selected and it was
found that 44 of them heat with gas. Perform the appropriate test of
hypothesis to determine whether the proportion of home heated by gas has
changed. Test using the 5% level of significance.

6) A class had a mean of 67 in the final exam and a standard deviation of
5. The class has more poorly performing students than those who perform
well. What can you say about the percentage of students who have scores
between 57 and 77?

7) Tough Road Tire Company wants to test its new brand of tires. The
engineering department of the company claims that the new tires last more
than 40,000 miles in the average. A sample of 40 tires was tested and found
that the average number of miles they lasted for the sample was 43,200 miles
with a standard deviation of 8000 miles. At the level of significance 0.05,
can the company claim that the tires last an average of more than 40,000
miles? Show all your work.

8) A weight loss program measured the weights lost by 12 participants. For
this sample of 12 participants the mean weights lost was 15.2 pounds and the
standard deviation was 5.3 pounds. Find the 95% confidence interval for the
average ( mean ) weight lost by the participants of this program. Assume
that the weights lost by the participants follow a normal distribution.

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Solution Summary

The solution provides step by step method for the calculation of normal probability and confidence intervals . Formula for the calculation and Interpretations of the results are also included.

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