Question

The grades on the last history exam had a mean of 80%. Assume the population of grades on history exams is known to be normally distributed with a standard deviation of 5. What percent of students earn a score between 70% and 80%?

Answer 1

Since we are given the mean μ = .80, and standard deviation σ = 5, we compute first the corresponding z scores of x₁ = .70 and x₂ = .80.
The formula for z score is z = (x-μ)/σ.

z₁ = (.70-.80)/5 = -0.02
z₂ = (.80-.80)/5 = 0

Then, using a z table, we find the probability of the corresponding z scores.
For z₁, it is 0.4920 and for z₂ it is 0.5000. (If you notice, x=μ, its probability will always be 0.5000).

We then find the difference between the two probabilities in order to find the percentage of students that earned a score between 70% and 80%
0.5000 - 0.4920 = 0.008 = 0.8%

Only 0.8% of students earned a score between 70% and 80%.

Answer 2

Answer:
`47.72\%`

Explanation:

We assume the population of grades on history exams is known to be normally distributed.

Given : Population mean :
`\mu=80\%=0.8`

Standard deviation :
`5\%=0.05`

Let x be the random variable that represents the grades on history exams .

Z-score :
`z=(x-\mu)/(\sigma)`

For x= 0.70

`z=(0.70-0.80)/(0.05)=-2`

For x=0.80

`z=(0.80-0.80)/(0.05)=0`

By using the standard normal distribution table , the probability of students earn a score between 70% and 80% will be :_

`P(-2<z<0)=P(z<0)-P(z<-2)\\\\=0.5- 0.0227501=0.477249\approx47.72\%`