Answer: Assume that I don't have IQ scores that are normally distributed with a mean of 105 in a standard deviation of 15.
find the probability that a randomly selected adult has an IQ between 95 and 115.
Solution:
Mean, μ = 105
Standard deviation, σ = 15
P(95 < X < 115) = P(95 - 105 /15 < Z < 115 - 105 / 15)
P(95 < X < 115) = P(- 0.67 < Z < 0.67)
P(95 < X < 115) = P(Z < 0.67) - P(Z < - 0.67)
P(95 < X < 115) = 0.7486 - 0.2514 (from z table)
P(95 < X < 115) = 0.4972
Therefore, the probability that a randomly selected adult has an IQ between 95 and 115 is 0.4972.
Get Answers For Free
Most questions answered within 1 hours.