Answer to Question #350587 in Statistics and Probability for sergu

Question

8. The average drive to work is 9.6 miles. Assume standard deviation is 1.8. If random sample of 36 employed who derive to work are selected, find the probability that mean of sample miles driven to work is between 9 and 10 miles.

Answer 1

Solution:

Let's denote given values:

$\mu=9.6 miles$ - population mean;

$\sigma=1.8 miles$ -standard deviation;

$n=36$ - sample number;

$X_1=9miles;$ $X_2=10miles$ -mean samples.

Find $z:$

$z_1=\frac{X_1-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{9-9.6}{\frac{1.8}{\sqrt{36}}}=-2;$

$z_1= -2;$ so, probability from z table $0.028$ , it means $2.8\%$ of X less than 9;

Find $z_2:$

$z_2=\frac{X_2-\mu}{\frac{\sigma}{\sqrt{n}}}=\frac{10-9.6}{\frac{1.8}{\sqrt{36}}}=1.33;$

From z table: $z_2=1.33,$ so probability $0.0918$. $1-0.918=0.9082;$ $90.82\%$ of X less than 10. So,

$P(9<X<10)=0.9082-0.028=0.8802;$

Answer:

$P(9<X<10)=88.02\%.$