Question

scores from a questionnaire measuring social anxiety form a normal distribution with a mean of μ = 50 and a standard deviation of σ = 10. what is the probability of obtaining a sample mean greater than m = 53. for a random sample of n = 4 people? σm = select z = select p = select for a random sample of n = 16 people? σm = select z = select p = select for a random sample of n = 25 people? σm = select z = select p = select

Explanation:

Step1: Calculate standard error for sample mean ($\sigma_M$)

The formula for the standard error of the mean is $\sigma_M=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard - deviation and $n$ is the sample size.

For $n = 4$:

$\sigma_M=\frac{10}{\sqrt{4}}=\frac{10}{2}=5$

For $n = 16$:

$\sigma_M=\frac{10}{\sqrt{16}}=\frac{10}{4}=2.5$

For $n = 25$:

$\sigma_M=\frac{10}{\sqrt{25}}=\frac{10}{5}=2$

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{M-\mu}{\sigma_M}$, where $M$ is the sample mean, $\mu$ is the population mean, and $\sigma_M$ is the standard error of the mean. Here, $\mu = 50$ and $M = 53$.

For $\sigma_M = 5$ (when $n = 4$):

$z=\frac{53 - 50}{5}=\frac{3}{5}=0.6$

For $\sigma_M = 2.5$ (when $n = 16$):

$z=\frac{53 - 50}{2.5}=\frac{3}{2.5}=1.2$

For $\sigma_M = 2$ (when $n = 25$):

$z=\frac{53 - 50}{2}=\frac{3}{2}=1.5$

Step3: Calculate the probability

We want $P(M>53)$, which is equivalent to $P(Z > z)$ in the standard normal distribution. Using the property $P(Z>z)=1 - P(Z\leq z)$. We can look up $P(Z\leq z)$ in the standard - normal table.

For $z = 0.6$:

$P(Z>0.6)=1 - P(Z\leq0.6)=1 - 0.7257 = 0.2743$

For $z = 1.2$:

$P(Z>1.2)=1 - P(Z\leq1.2)=1 - 0.8849 = 0.1151$

For $z = 1.5$:

$P(Z>1.5)=1 - P(Z\leq1.5)=1 - 0.9332 = 0.0668$

Answer:

For $n = 4$:
$\sigma_M = 5$, $z = 0.6$, $p = 0.2743$
For $n = 16$:
$\sigma_M = 2.5$, $z = 1.2$, $p = 0.1151$
For $n = 25$:
$\sigma_M = 2$, $z = 1.5$, $p = 0.0668$