Question

1. scores on a standardized reading test for fourth - grade students form a normal distribution with $mu=71$ and $sigma = 24$. what is the probability of obtaining a sample mean greater than $m = 63$ for each of the following?

a. a sample of $n = 9$ students
b. a sample of $n = 36$ students
c. a sample of $n = 64$ students

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{M}=\frac{\sigma}{\sqrt{n}}$.
For part a: $n = 9$, $\sigma = 24$, so $\sigma_{M}=\frac{24}{\sqrt{9}}=\frac{24}{3}=8$.
For part b: $n = 36$, $\sigma = 24$, so $\sigma_{M}=\frac{24}{\sqrt{36}}=\frac{24}{6}=4$.
For part c: $n = 64$, $\sigma = 24$, so $\sigma_{M}=\frac{24}{\sqrt{64}}=\frac{24}{8}=3$.

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{M-\mu}{\sigma_{M}}$.
For part a: $M = 63$, $\mu = 71$, $\sigma_{M}=8$, so $z=\frac{63 - 71}{8}=\frac{-8}{8}=-1$.
For part b: $M = 63$, $\mu = 71$, $\sigma_{M}=4$, so $z=\frac{63 - 71}{4}=\frac{-8}{4}=-2$.
For part c: $M = 63$, $\mu = 71$, $\sigma_{M}=3$, so $z=\frac{63 - 71}{3}=\frac{-8}{3}\approx - 2.67$.

Step3: Find the probability

We want $P(M>63)$, which is equivalent to $P(z>-z_{0})$ (where $z_{0}$ is the calculated negative z - score). Since $P(z > - z_{0})=1 - P(z\leq - z_{0})$.
Using the standard normal table:
For part a: $P(z>-1)=1 - P(z\leq - 1)$. From the standard normal table, $P(z\leq - 1)=0.1587$, so $P(z>-1)=1 - 0.1587 = 0.8413$.
For part b: $P(z>-2)=1 - P(z\leq - 2)$. From the standard normal table, $P(z\leq - 2)=0.0228$, so $P(z>-2)=1 - 0.0228 = 0.9772$.
For part c: $P(z>-2.67)=1 - P(z\leq - 2.67)$. From the standard normal table, $P(z\leq - 2.67)=0.0038$, so $P(z>-2.67)=1 - 0.0038 = 0.9962$.

Answer:

a. $0.8413$
b. $0.9772$
c. $0.9962$