Question

A set of biology exam scores are normally distributed with a mean of 707070 points and a standard deviation of 666 points. Let XXX represent the score on a randomly selected exam from this set. Find P(64

Answer 1

Answer:

$P(64<x<76)=0.68$

Step-by-step explanation:

Given that, the mean, $\mu$, and the standard deviation, $\sigma$, of normally distributed exam score are :

$\mu=70$

$\sigma=6$

Let $x$ 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

$\frac{64-70}{6}<z<\frac{76-70}{6}= -1<z<1$

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, $P(64<x<76)= P(x<76)-P(x<64)$

$\Rightarrow P(64<x<76)=0.84134-0.15866=0.68268.$

Hence, $P(64<x<76)=0.68$