# A population of values has a normal distribution with \mu=99.6 and \sigma=35.1.

A population of values has a normal distribution with $\mu =99.6$ and $\sigma =35.1$. You intend to draw a random sample of size $n=84$.
Find the probability that a single randomly selected value is between 98.5 and 100.7.
$P\left(98.5?

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Delorenzoz

Step 1
From the provided information,
Mean $\left(\mu \right)=99.6$
Standard deviation $\left(\sigma \right)=35.1$
$X\sim N\left(99.6,35.1\right)$
Sample size $\left(n\right)=84$
Step 2
The required probability that a single randomly selected value is between 98.5 and 100.7 can be obtained as:
$P\left(98.5
$=P\left(-0.031
$=P\left(Z<0.031\right)-P\left(Z<-0.031\right)$
$=P\left(Z<0.031\right)-\left[1-P\left(z<0.031\right)\right]$
$=2P\left(Z<0.031\right)-1$
$=2\left(0.5124\right)-1=0.0248$ (Using standard normal table)
Thus, the required probability is 0.0248.

Jeffrey Jordon