Answer: 21
Explanation:
Given : The scores on the midterm exam are normally distributed with
![\mu=72.3\\\\\sigma=8.9](https://img.qammunity.org/2020/formulas/mathematics/college/3o9ktqrwkb27di9yjkuqczxjk2ztkhkbx7.png)
Let X be random variable that represents the score of the students.
z-score:
![z=(x-\mu)/(\sigma)](https://img.qammunity.org/2020/formulas/mathematics/high-school/10fia1p0qwvlz4zhb867kzy3u7bscognwz.png)
For x=82
![z=(82-72.3)/(8.9)\approx1.09](https://img.qammunity.org/2020/formulas/mathematics/college/3cq41ybxbv1yerz97xq88likdsbmn2beo8.png)
For x=90
![z=(90-72.3)/(8.9)\approx1.99](https://img.qammunity.org/2020/formulas/mathematics/college/i9lntms759mbswqhmr1bkmgteqix54qf51.png)
Now, the probability of the students in the class receive a score between 82 and 90 ( by using standard normal distribution table ) :-
![P(82<X<90)=P(1.09<z<1.99)\\\\=P(z<1.99)-P(z<1.09)\\\\=0.9767-0.8621=0.1146](https://img.qammunity.org/2020/formulas/mathematics/college/woywgao0fxo54tw1zbhnp0c15iitwnn0yn.png)
Now ,the number of students expected to receive a score between 82 and 90 are :-
![184*0.1146=21.0864\approx21](https://img.qammunity.org/2020/formulas/mathematics/college/uom4udoflqmvau8fdht3qm7h33omwbe524.png)
Hence, 21 students are expected to receive a score between 82 and 90 .