A population of values has a normal distribution with \mu = 181 and \sigma = 41.8. You intend to draw a

Question

A population of values has a normal distribution with $μ = 181$ and $σ = 41.8$ . You intend to draw a random sample of size $n = 144$ .
Find the probability that a single randomly selected value is between 176.5 and 183.1.
$P (176.5 < M < 183.1) =$ ?
Write your answers as numbers accurate to 4 decimal places.

Answer 1

Given that,
$μ = 181$ ,
$σ = 41.8$
$n = 144$
The probability that a sample size, $n = 144$ is randomly selected with a mean between 176.5 and 183.1 is,
$P (176.5 < \overset{―}{X} < 183.1) = P (\frac{176.5 - 181}{\frac{41.8}{\sqrt{144}}} < \frac{\overset{―}{X} - μ}{\frac{σ}{\sqrt{144}}} < \frac{183.1 - 181}{\frac{41.8}{\sqrt{144}}})$
$= P (- 1.292 < z < 0.603)$
$= P (z < 0.603) - P (z < - 1.292)$
$= {\begin{matrix} (= N O R M D I S T (0.050)) \\ - (= N O R M D I S T (- 0.108)) \end{matrix}} (\begin{matrix} (U s i n g E x c e l f u n c t i o n, \\ (= N O R M D I S T (z)) \end{matrix})$
$= 0.7267 - 0.0982 = 0.6285$
Therefore, the required probability is, 0.6285.