Question

μ = 277
σ = 32
a) if you take a sample of size 14, what can you say what the shape of the sampling distribution for the sample mean?
the sampling distribution will be approximately normal
because: (check all that apply)
population is normal
n is less than 30
population is not normal
n is at least 30
b) for a sample of size 14, state the mean and the standard deviation of the sampling distribution of the sample mean.
μ =
σ =
round to 2 decimals
c) find the probability of getting a sample mean greater than 288.89 when taking a sample of size 14 from this population.
p(ˉx > 288.89) =
round to 2 decimals

Explanation:

Step1: Recall sampling - distribution properties

If the population is normal, the sampling distribution of the sample - mean is normal regardless of the sample size. Given $\mu = 277$ and $\sigma=32$, and $n = 14$.

Step2: Calculate the mean of the sampling distribution of the sample - mean

The mean of the sampling distribution of the sample - mean $\mu_{\bar{x}}$ is equal to the population mean $\mu$. So, $\mu_{\bar{x}}=\mu = 277$.

Step3: Calculate the standard deviation of the sampling distribution of the sample - mean

The standard deviation of the sampling distribution of the sample - mean (also known as the standard error) is given by $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Substituting $\sigma = 32$ and $n = 14$, we have $\sigma_{\bar{x}}=\frac{32}{\sqrt{14}}\approx\frac{32}{3.742}\approx8.55$.

Step4: Calculate the z - score

The z - score is calculated using the formula $z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}$. We want to find $P(\bar{X}>288.89)$. First, calculate the z - score: $z=\frac{288.89 - 277}{8.55}=\frac{11.89}{8.55}\approx1.39$.

Step5: Find the probability

We know that $P(\bar{X}>288.89)=P(Z > 1.39)$. Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard normal table $P(Z\leq1.39) = 0.9177$. So $P(Z > 1.39)=1 - 0.9177 = 0.0823\approx0.08$.

Answer:

b) Mean of the sampling distribution of the sample - mean: $\mu_{\bar{x}} = 277$
Standard deviation of the sampling distribution of the sample - mean: $\sigma_{\bar{x}}\approx8.55$
c) $P(\bar{X}>288.89)\approx0.08$