Answer:
0.8413 or 84.13%
Step-by-step explanation:
Given : The mean is 72 inches and the standard deviation is 15 inches
To Find : What is the probability that in a randomly selected year, the snowfall was less than 87 inches
Solution:
Mean = ![\mu = 72](https://tex.z-dn.net/?f=%5Cmu%20%3D%2072)
Standard deviation = ![\sigma = 15](https://tex.z-dn.net/?f=%5Csigma%20%3D%2015)
Formula : ![z=\frac{x-\mu}{\sigma}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-%5Cmu%7D%7B%5Csigma%7D)
We are supposed to find the probability that in a randomly selected year, the snowfall was less than 87 inches
So, x = 87
Substitute the values in the formula
![z=\frac{87-72}{15}](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B87-72%7D%7B15%7D)
![z=1](https://tex.z-dn.net/?f=z%3D1)
Now to find P(z<87) refer the z table
P(Z<87)=0.8413 = 84.13%
So, the probability that in a randomly selected year, the snowfall was less than 87 inches if the mean is 72 inches and the standard deviation is 15 inches is 0.8413 or 84.13%