Question

6.5 sampling distribution of the sample mean and the central limit theorem. compute probabilities of a sample mean
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a population of values has a normal distribution with $mu = 6.9$ and $sigma = 39.4$. you intend to draw a random sample of size $n = 117$. please show your answers as numbers accurate to 4 decimal places.
find the probability that a single randomly selected value is less than 10.2.
$p(x < 10.2)=.5334$
find the probability that a sample of size $n = 117$ is randomly selected with a mean less than 10.2.
$p(\bar{x}< 10.2)=$
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Explanation:

Step1: Calculate the standard error

The standard error of the mean is given by $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 39.4$ and $n = 117$. So, $\sigma_{\bar{x}}=\frac{39.4}{\sqrt{117}}\approx3.6378$.

Step2: Calculate the z - score

The z - score formula for the sample mean is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. Here, $\bar{x}=10.2$, $\mu = 6.9$, and $\sigma_{\bar{x}}\approx3.6378$. So, $z=\frac{10.2 - 6.9}{3.6378}=\frac{3.3}{3.6378}\approx0.9071$.

Step3: Find the probability

We want to find $P(\bar{X}<10.2)$, which is equivalent to $P(Z < 0.9071)$ using the standard normal distribution. Looking up the value in the standard - normal table (or using a calculator with a normal - distribution function), $P(Z < 0.9071)\approx0.8186$.

Answer:

$0.8186$