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
Answer:
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!
