Given:
Let the random variable x is normally distributed with mean
and
![\sigma=5](https://img.qammunity.org/2021/formulas/mathematics/college/fi7x4iw6w6e3kmti5t58qi4i1fu2huzho6.png)
We need to determine the probability of
![P(X<80)](https://img.qammunity.org/2021/formulas/mathematics/college/vkjo4fa5r7hr4pugnqkv53dl19k8f6p7ei.png)
Probability of
:
The formula to determine the value of
is given by
![Z=(X-\mu)/(\sigma)](https://img.qammunity.org/2021/formulas/mathematics/high-school/xjryp8ydeqt32yspvlaxie9csx43qbw8sp.png)
Thus, we have;
![P(X<80)=P(Z<(80-90)/(5))](https://img.qammunity.org/2021/formulas/mathematics/middle-school/96n0ctsoendy1w8q594vcimq672ivwmgyt.png)
Simplifying, we get;
![P(X<80)=P(Z<(-10)/(5))](https://img.qammunity.org/2021/formulas/mathematics/middle-school/wfgiclhx7a6dq4rrlimdvfr1wvab98yh9u.png)
![P(X<80)=P(Z<-2)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/st5ffo1v519ll5i5fz2idium57e61pnkwi.png)
Using the normal distribution table, the value of -2 is given by 0.0228
![P(X<80)=0.0228](https://img.qammunity.org/2021/formulas/mathematics/middle-school/byej7yw0lg3c0owvk4mharnj1hrgfyzi3z.png)
Thus, the value of
is 0.0228