Almost all medical schools require applicants to take the Medical College Admission Test (MCAT). To estimate the mean score of those who took the MCAT at WSSU, you will obtain the scores of an SRS of students. The scores follow a Normal distribution, and from published information you know that the standard deviation is 6.5. Suppose that (unknown to you) the mean score of those taking the MCAT at WSSU is 25.0. You sample 25 students. What is the standard deviation of the sampling distribution of x-bar?

1.3

Step-by-step explanation:

The given parameters are;

The standard deviation, σ = 6.5

The mean score, = 25.0

The number of students in the sample, n = 25 students

Whereby the standard deviation of the sampling distribution of is the standard deviation of , we have;

The standard deviation of = σ/√(n)

∴ The standard deviation of = 6.5/√(25) = 1.3

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step-by-step explanation:

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a triangle is base times height divided by two. so multiply 7.9 and 5 and then divide it by 2 to get 19.75inches.

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