Question

5- A national test is held to admit students to the public universities. The scores on this test are normally distributed with a mean of 350 and a standard deviation of 60. Mary knows if she wants to be admitted to a certain university, her score should be better than at least 85% of the students who took the test. a) (6%) Mary takes the test and scores 435. Will she be admitted to that certain university? b) (79) If a random sample of 40 students selected, what is the probability that the sample average of test scores is at most 370? c) (7%) How large a sample size would be required to ensure that, the probability of having the average of 375 for test scores is at least 0.95?

Answer 1

Given that the test scores are Normally distributed with
u 350
and $\sigma=60$ .

Given that to get into a certain university, her score has to be better than at least 85% of the students who took the test.

ie the the score for which P(X>a)=0.15. The point a will be the minimum score From standard Normal table, we get the value for Z as 1.0364.

We know that $Z=\frac{X-\mu}{\sigma}$

X – 350 60 1.0364 X = 350 +1.0364 * 60 = 412.184

a. Mary scored 435. Since her score 435>412, the 85th percentile, she will be admitted to certain University.

b. We know that sample mean follows a Normal with mean
u 350
and sd =$\frac{\sigma}{\sqrt{n}}$. We are given that n=40, the probability that the sample mean score is at most 370 is

$P(\bar{x}<370)$.

We know that

I N - g/n

When $\bar{x}=370,$ $Z=\frac{370-350}{60/\sqrt{40}}$

  $Z=\frac{20}{9.4868}=2.1082$

$P(\bar{x}<370)=P(Z<2.1082)=0.9825$

Therefore, the probability that the sample mean score is at most 370 is 0.9825

(c). To find the sample size so that $P(\bar{x}>375)\ge0.95$

The Z value corresponding to 0.95 is 1.6449.

$\frac{370-350}{60/\sqrt{n}}=1.6449$

$\sqrt{n}*\frac{20}{60}=1.6449$

$\sqrt{n}=\frac{1.6449*60}{20}$

$\sqrt{n}=4.9347$

Squaring on both sides, we get

$n=24.35\sim 25$.

The sample size has to be at least 25.