Answer: 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
