Answer:
![P(64](https://tex.z-dn.net/?f=P%2864%3Cx%3C76%29%3D0.68)
Step-by-step explanation:
Given that, the mean,
, and the standard deviation,
, of normally distributed exam score are :
![\mu=70](https://tex.z-dn.net/?f=%5Cmu%3D70)
![\sigma=6](https://tex.z-dn.net/?f=%5Csigma%3D6)
Let
represent the score on a randomly selected exam from this set
The z-score, for the randomly selected exam score x:
For 64<x<76, the range of z-score is
![\frac{64-70}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B64-70%7D%7B6%7D%3Cz%3C%5Cfrac%7B76-70%7D%7B6%7D%3D%20-1%3Cz%3C1)
Now, from the z-score table, the value for z=-1 is the probability for x<64.
As the value for z=-1 is 0.15866, so
P(x<64)=0.15866.
Similarly, the value for z=0.67 is 0.84134, so
P(x<76)=0.84134
So, ![P(64](https://tex.z-dn.net/?f=P%2864%3Cx%3C76%29%3D%20P%28x%3C76%29-P%28x%3C64%29%20%5C%5C%5C%5C)
![\Rightarrow P(64](https://tex.z-dn.net/?f=%5CRightarrow%20P%2864%3Cx%3C76%29%3D0.84134-0.15866%3D0.68268.)
Hence, ![P(64](https://tex.z-dn.net/?f=P%2864%3Cx%3C76%29%3D0.68)