If 142 adults are randomly selected, what is the probability that the sample mean would differ from the population mean by more than 1.5 kilograms

Complete question;

The mean weight of an adult is 76 kilograms with a variance of 100.

If 142 adults are randomly selected, what is the probability that the sample mean would differ from the population mean by more than 1.5 kilograms

7.34% 0r 0.0734

Step-by-step explanation:

We have mean u = 76

Standard deviation = square root of variance

Sd = √100 = 10

N = 142

S = 10/√142

= 10/11.92

= 0.84

X is going to differ by more than 1.5 or less than 1.5

76-1.5 = 74.5

76+1.5 = 77.5

At x = 74.5

Z = (74.5-76)/0.84 = -1.79

At x = 77.5

Z = (77.5-76)/0.8 = 1.79

P value of z at -1.79 = 0.036

P value of z at 1.79 = 0.9633

0.9633-0.0367 = 0.9266

Which is 92.66% probability of differing by 1.5

Probability it differs by more:

P+92.66 = 100

P = 100-92.66

= 7.34% or 0.0734

Thank you!