Question

Assume that adults have it scores that are normally distributed with a mean of 100 standard deviation of 15 find probability that randomly selected adult has an Iq between 89 and 111

Answer 1

Answer:

0.5346

Step-by-step explanation:

Find the z-scores.

z = (x − μ) / σ

z₁ = (89 − 100) / 15

z₁ = -0.73

z₂ = (111 − 100) / 15

z₂ = 0.73

Find the probability.

P(-0.73 < Z < 0.73)

= P(Z < 0.73) − P(Z < -0.73)

= 0.7673 − 0.2327

= 0.5346