A professor has given the same exam to every Stats class for the last 30 years (during which thousands of students have taken the exam), with a historical mean of 70.8 and a standard deviation of 10.1. Moreover, the historical exam results closely follow a normal distribution. This year he experiments by using a totally new teaching method to the class of 25 students, but he gives them the same exam as in all prior years. Part a How likely is a randoml

A. 18.14%

B. The scores have to be between 66.84 and 74.76 with 95% certainty.

Correct statement and question:

A professor has given the same exam to every Stats class for the last 30 years (during which thousands of students have taken the exam), with a historical mean of 70.8 and a standard deviation of 10.1. Moreover, the historical exam results closely follow a normal distribution. This year he experiments by using a totally new teaching method to the class of 25 students, but he gives them the same exam as in all prior years.

A. How likely is a randomly selected student from last year before the new teaching method to have scored more than 80?

B. The professor considers his new teaching method to be successful if, in his own words, "he sees more scores around the mean". What exam results are required for this to happen (with 95% certainty)?

Source:

Previous question that can be found at coursehero

Step-by-step explanation:

Part A.

μ = 70.8

σ = 10.1

P (x > 80) = 1 - P (x - μ/σ > 80 - μ/σ)

P (x > 80) = 1 - P (z > 80 - 70.8/10.1)

P (x > 80) = 1 - P (z > 0.9109)

P (x > 80) = 1 - 0.8186 (Using our z-table we use the value for z-score = 0.91)

P (x > 80) = 0.1814

Part B.

Let's recall that the formula for calculating the 95% confidence interval is:

μ +/- 1.96 * σ/√n

σ/√n = standard error

This formula of standard error is valid if:  

Replacing with the values we have:

70.8 +/- 1.96 * 10.1/√25

70.8 +/- 1.96 *2.02

70.8 +/- 3.96

Lower limit = 66.84

Upper limit = 74.76

The scores have to be between 66.84 and 74.76 with 95% certainty.

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