In a large population of college-educated adults, the mean iq is 118 with a standard deviation of 20. suppose 200 adults from this population are randomly selected for a market research campaign. the probability that the sample mean iq is greater than 120 is

0.0793

Step-by-step explanation:

Let the IQ of the educated adults be X then;

Assume X follows a normal distribution with mean 118 and standard deviation of 20.

This is a sampling question with sample size, n =200

To find the probability that the sample mean IQ is greater than 120:

P(X > 120) = 1 - P(X < 120)

Standardize the mean IQ using the sampling formula : Z = (X - μ) / σ/sqrt n

Where;  X = sample mean IQ; μ =population mean IQ; σ = population standard deviation and n = sample size

Therefore, P(X>120) = 1 - P(Z < (120 - 118)/20/sqrt 200)

                                 = 1 - P(Z< 1.41)

The P(Z<1.41) can then be obtained from the Z tables and the value is 0.9207

Thus; P(X< 120) = 1 - 0.9207

                          = 0.0793