Student scores on exams given by a certain instructor have mean 74 and standard deviation 14. This instructor is about to give two exams, one to a class of size 25 and the other to a class of size 64.

Required:
a. Approximate the probability that the average test score in the class of size 25 exceeds 80.
b. Repeat part (a) for the class of size 64.
c. Approximate the probability that the average test score in the larger class exceeds that of the other class by over 2.2 points.

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tallinn tallinn · Answer 1 · 2021-07-18T12:30:09+02:00

Answer:

a) Hence the probability that the average test score in the class of size 25 exceeds 80.

P ( X > 74) = 0.9838

P ( Z > 2.14) = 0.0162

b) Hence the probability that the average test score for the class of size 64

P ( X > 74) = 0.9838

P ( Z > 2.14) = 0.0003

c) Probability of the difference exceeding 2.2 = 0.9936

P (Z < 2.49) = 0.0064

Step-by-step explanation:

Let's assume a normal distribution.

Now,

a) For a class of 25

[tex]P ( X > 74) = P (Z > 80 - 74/ 14\sqrt(25) )\\\\= P (Z < 6 / 2.8)\\\\= P ( Z < 2.14)\\= 0.9838\\[/tex]

[tex]P ( Z > 2.14) = 1- 0.9838\\= 0.0162[/tex]

b)

Similarly:

For the class of 64

[tex]P ( X > 74) = P (Z > 80 - 74/ 14\sqrt(64) )\\\\= P (Z < 6 / 1.75)= P ( Z < 3.428)\\= 0.9838\\[/tex]

[tex]P ( Z > 2.14) = 1- 0.9997\\= 0.0003[/tex]

c) Probability of the difference exceeding 2.2

[tex]= P (Z > 2.2/\sqrt{14 * {(1/25) + (1/64}\\[/tex]

[tex]= P (Z > 2.49)\\= 0.9936[/tex]

P (Z < 2.49)

= 1 - 0.9936

= 0.0064