Question

1. a normal population has a mean μ = 33 and standard deviation σ = 9. what is the probability that a randomly chosen value will be greater than 44? \\(\frac{44 - 33}{9} = 1.22\\)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x = 44$, $\mu=33$, and $\sigma = 9$.
Substitute the values into the formula: $z=\frac{44 - 33}{9}=\frac{11}{9}\approx1.22$.

Step2: Find the probability

We want to find $P(X>44)$, which is equivalent to $P(Z > 1.22)$ in the standard normal distribution.
We know that $P(Z>z)=1 - P(Z\leq z)$. From the standard normal table, $P(Z\leq1.22) = 0.8888$.
So $P(Z > 1.22)=1 - 0.8888 = 0.1112$.

Answer:

The probability that a randomly chosen value will be greater than 44 is approximately 0.1112 (or 11.12%).