A population of values has a normal distribution with \mu=26.8 and \sigma=33.8. Yo

Question

A population of values has a normal distribution with $μ = 26.8$ and $σ = 33.8$ . You intend to draw a random sample of size $n = 89$ .
Find the probability that a sample of size $n = 89$ is randomly selected with a mean between 17.1 and 25.
$P (17.1 < M < 25) =$ ?
Write your answers as numbers accurate to 4 decimal places.

Answer 1

Step 1
It is given that a population of values distributed normally with mean 26.8 and standard deviation 33.8. The sample size is 89.
Step 2
Calculate the probability that a sample of size $n = 89$ is randomly selected with a mean between 17.1 and 25 is as follows:
$P (17.1 < X < 25) = P (\frac{17.1 - μ}{\frac{σ}{\sqrt{n}}} < \frac{X - μ}{\frac{σ}{\sqrt{n}}} < \frac{25 - μ}{\frac{σ}{\sqrt{n}}})$
$= P (\frac{17.1 - 26.8}{\frac{33.8}{\sqrt{89}}} < Z < \frac{25 - 26.8}{\frac{33.8}{\sqrt{89}}})$
$P (- \frac{9.7}{3.58279} < Z < \frac{1.8}{3.58279})$
$= P (- 2.707 < Z < - 0.502)$
$= P (Z < - 0.053) - P (Z < - 0.287) (\begin{matrix} U s i n g E x c e l f u n c t i o n s, \\ P (Z < - 0.053) = 0.4789 \\ P (Z < - 0.287) = 0.3871 \end{matrix})$
$= 0.3078 - 0.0034 = 0.3044$
Therefore, the value of $P (17.1 < M < 25) = 0.3044$ .

A population of values has a normal distribution with \mu=26.8 and \sigma=33.8. Yo

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