Almost all medical schools in the United States require stud

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate the mean score of those who took the MCAT on your campus, 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 6.3. Suppose that (unknown to you) the mean score of those taking the MCAT on your campus is 26.

In answering the following, use z-scores rounded to two decimal places.

If you choose one student at random, what is the probability (±0.0001) that the student\'s score is between 20 and 30?

You sample 25 students. What is the standard deviation (±0.01) of sampling distribution of their average score x?

What is the probability (±0.0001) that the mean score of your sample is between 20 and 30?

Solution

A)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    20      
x2 = upper bound =    30      
u = mean =    26      
          
s = standard deviation =    6.3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.952380952      
z2 = upper z score = (x2 - u) / s =    0.634920635      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.170451908      
P(z < z2) =    0.737259911      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.566808003   [answer]

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b)

standard error of the mean = s / sqrt(n) = 6.3/sqrt(25) = 1.26 [answer]

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c)

We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as          
x1 = lower bound =    20      
x2 = upper bound =    30      
u = mean =    26      
n = sample size =    25      
s = standard deviation =    6.3      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u) * sqrt(n) / s =    -4.761904762      
z2 = upper z score = (x2 - u) * sqrt(n) / s =    3.174603175      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    9.58871*10^-7      
P(z < z2) =    0.999249792      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.999248833   [ANSWER]