Answer to Question #347568 in Statistics and Probability for John

Question

A normally distributed population has a mean of 1,214 and a standard deviation of 122. Find the probability that a single randomly selected element X of the population is between 1,100 and 1,300.

Answer 1

$P(1,100<X<1,300)=P(z_1<z<z_2),$

where z₁ and z₂ are z-scores of 1,100 and 1,300.

Let's find them.

$z=\frac{x-\mu}{\sigma};$ $\mu=1,214;$ $\sigma=122.$

$z_1=\frac{1,100-1,214}{122}\approx-0.93,$

$z_2=\frac{1,300-1,214}{122}\approx0.70.$

So we have to rewrite the probability ang use z-table:

$P(1,100<X<1,300)=P(-0.93<z<0.70)=$

$=0.7580-0.1762=0.5818=58.18\%.$

Answer: 58.18%.