Question

suppose that the distribution of scores on the graduate record exam (gre) is approximately normal, with a mean of $mu = 150$ and a standard deviation of $sigma = 5$. for the population of students who have taken the gre: a. what proportion have gre scores less than 145? b. what proportion have gre scores greater than 157? c. what is the minimum gre score needed to be in the highest 20% of the population? d. if a graduate school accepts only students from the top 10% of the gre distribution, what is the mini - mum gre score needed to be accepted?

Explanation:

Step1: Standardize the score

We use the z - score formula $z=\frac{x-\mu}{\sigma}$, where $\mu = 150$ is the mean and $\sigma = 5$ is the standard deviation.

Step2: Calculate proportion for part a

For $x = 145$, $z=\frac{145 - 150}{5}=\frac{- 5}{5}=-1$. Looking up the z - value in the standard normal table, $P(Z < - 1)=0.1587$.

Step3: Calculate proportion for part b

For $x = 157$, $z=\frac{157-150}{5}=\frac{7}{5} = 1.4$. Then $P(Z>1.4)=1 - P(Z < 1.4)$. From the standard - normal table, $P(Z < 1.4)=0.9192$, so $P(Z>1.4)=1 - 0.9192 = 0.0808$.

Step4: Find z - value for part c

If we want to be in the highest 20% of the population, we want the lower 80% of the population. Looking up in the standard - normal table, the z - value corresponding to a cumulative probability of 0.8 is approximately $z = 0.84$. Then, using the z - score formula $z=\frac{x-\mu}{\sigma}$, we solve for $x$: $0.84=\frac{x - 150}{5}$, so $x=150+0.84\times5=150 + 4.2=154.2$.

Step5: Find z - value for part d

If we want to be in the top 10% of the population, we want the lower 90% of the population. Looking up in the standard - normal table, the z - value corresponding to a cumulative probability of 0.9 is approximately $z = 1.28$. Then, using the z - score formula $z=\frac{x-\mu}{\sigma}$, we solve for $x$: $1.28=\frac{x - 150}{5}$, so $x=150+1.28\times5=150 + 6.4=156.4$.

Answer:

a. 0.1587
b. 0.0808
c. 154.2
d. 156.4