If the heights of 300 students are normally distributed with mean 68…

The correct answer is Option (1) → (A) and (B) only

Standardising with \( Z=\frac{X-\mu}{\sigma}=\frac{X-68}{3} \):

    \( \displaystyle X=72:\ Z=\frac{4}{3}\approx 1.33 \)

    \( \displaystyle X=64:\ Z=-1.33 \)
    \( \displaystyle X=65:\ Z=-1 \)
    \( \displaystyle X=71:\ Z=1 \)

    \( \displaystyle P(X\gt 72)=P(Z\gt 1.33)=0.0918 \)

By the symmetry of the normal curve,

    \( \displaystyle P(X\le 64)=P(Z\le -1.33)=0.0918 \)

    \( \displaystyle 300\times 0.0918=27.54 \)

A number of students is a whole number, so 27.54 is taken as 27 students.

    \( \displaystyle P(65\le X\le 71)=P(-1\le Z\le 1) \)

    \( \displaystyle =2\times 0.3413=0.6826 \)

    \( \displaystyle 300\times 0.6826=204.78 \)

So 27 students are taller than 72 inches, 27 students are at most 64 inches tall, and about 205 students are between 65 and 71 inches.

Checking the statements

(A) There are 27 students whose height is greater than 72 inches. — Correct. \( 300\times 0.0918=27.54 \), that is, 27 students.

(B) The number of students having heights less than or equal to 64 inches is 27. — Correct. By symmetry the probability is again 0.0918, giving 27 students.

(C) 210 students have their heights between 65 and 71 inches. — Incorrect. \( 300\times 0.6826=204.78 \), about 205 students, not 210.

(D) There are 35 students whose height is greater than 72 inches. — Incorrect. \( 300\times 0.0918 \) gives 27 students, not 35.