The grades on the last history exam had a mean of 80%. Assume the population of grades on history exams is known to be distributed normally, with a standard deviation of 5%. Using the empirical rule, approximately what percent of students earned a score between 70% and 80%? 3.5% 5% 47.5% 95%

We are given that the grades on the last history exam had a mean of 80%. Assume the population of grades on history exams is known to be distributed normally, with a standard deviation of 5%.

Let X = grades on history exams

So, X ~ N([tex]\mu = 0.80 , \sigma^{2}=0.05^{2}[/tex])

The standard normal z score distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

So, percent of students earning a score between 70% and 80% is given by = P(0.70 < X < 0.80) = P(X < 0.80) - P(X <= 0.70)

P(X < 0.80) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{0.80-0.80}{0.05}[/tex] ) = P(Z < 0) = 0.50 {using z table}

P(X <= 0.70) = P( [tex]\frac{X-\mu}{\sigma}[/tex] <= [tex]\frac{0.70-0.80}{0.05}[/tex] ) = P(Z <= -2) = 1 - P(Z < 2) = 1 - 0.97725

= 0.02275

So, P(0.70 < X < 0.80) = 0.50 - 0.02275 = 0.47725 or 47.7%

Therefore, 47.7% of students earned a score between 70% and 80%.