Question

a population of values has a normal distribution with \\( \mu = 242.1 \\) and \\( \sigma = 57.7 \\). if a random sample of size \\( n = 24 \\) is selected,

a. find the probability that a single randomly selected value is greater than 253.9. round your answer to four decimals.
\\( p(x > 253.9) = \square \\)

b. find the probability that a sample of size \\( n = 24 \\) is randomly selected with a mean greater than 253.9. round your answer to four decimals.
\\( p(\bar{x} > 253.9) = 0.1582 \\)

Explanation:

Part (a)

Step 1: Calculate the z-score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$. Given that $\mu = 242.1$, $\sigma=57.7$ and $x = 253.9$.
Substitute the values into the formula: $z=\frac{253.9 - 242.1}{57.7}=\frac{11.8}{57.7}\approx0.2045$

Step 2: Find the probability $P(X>253.9)$

We know that $P(X > x)=1 - P(X\leq x)$. For a normal distribution, $P(X\leq x)=\Phi(z)$, where $\Phi(z)$ is the cumulative distribution function of the standard normal distribution.
So $P(X > 253.9)=1 - P(Z\leq0.2045)$. Looking up the value of $\Phi(0.2045)$ in the standard normal table (or using a calculator with normal distribution functions), $\Phi(0.2045)\approx0.5810$.
Then $P(X > 253.9)=1 - 0.5810 = 0.4190$

Answer:

For part (a), $P(X > 253.9)\approx\boldsymbol{0.4190}$