The average IQ score follows a Normal distribution, with a mean of μ = 100 and a standard deviation of σ = 15. What is the probability that the mean IQ score of 25 randomly selected people will be greater than 105? 0 0.0478 0.50 0.68 0.9522

0 0.0478

Step-by-step explanation:

Since the average IQ score follows a Normal distribution, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ/√n

Where

x = IQ scores

µ = mean score

σ = standard deviation

n = number of samples

From the information given,

µ = 100

σ = 15

n = 25

the probability that the mean IQ score of 25 randomly selected people will be greater than 105 is expressed as

P(x > 105) = 1 - P(x ≤ 105)

For x = 105,

z = (105- 100)/(15/√25) = 1.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.9525

P(x > 105) = 1 - 0.9525 = 0.0475