The summer income of the 3,408 students at Centennial High School last year was normally distributed with mean $1,751 and standard deviation $421.
Approximately what percent of the students had incomes between $1,000 and $3,000?
Round to the nearest percent? See lesson 1-4 Example 4 if necessary.
O 1.19%
99.85%
O 2.97%
0 27.61%
096.1%

You want to find

P(1000 < X < 3000)

where X is normally distributed with mean 1751 and standard deviation 421. Transform X to Z, so that it follows the standard normal distribution with mean 0 and standard deviation 1 using the relation

X = 1751 + 421Z   ==>   Z = (X - 1751)/421

Then

P(1000 < X < 3000) = P((1000 - 1751)/421 < (X - 1751)/421 < (3000 - 1751)/421)

… ≈ P(-1.783 < Z < 2.967)

… ≈ P(Z < 2.967) - P(Z < -1.783)

… ≈ 0.9985 - 0.0373

… ≈ 0.9612

so that approximately 96.1% of the students fall in this income range.