Question

question 4
the personal savings of the young saver club were normally distributed with a mean of $975 and a standard deviation of $88. what is the probability that a randomly selected saver has an account total between $1063 and $1151?
0.135
0.68
0.34
0.025

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set.
For $x = 1063$, $\mu=975$, and $\sigma = 88$, $z_1=\frac{1063 - 975}{88}=\frac{88}{88}=1$.
For $x = 1151$, $\mu = 975$, and $\sigma=88$, $z_2=\frac{1151 - 975}{88}=\frac{176}{88}=2$.

Step2: Use the standard normal distribution table

We want to find $P(1We know that $P(Z < 2)-P(Z < 1)$.
From the standard - normal table, $P(Z < 2)=0.9772$ and $P(Z < 1)=0.8413$.
So $P(1

Answer:

0.135