Question

Assume that IQ scores are normally distributed, with a standard deviation of 11 points and a mean of 100 points.
If 85 people are chosen at random, what is the probability that the sample mean of IQ scores will not differ from the population mean by more than 2 points?

Answer 1

Answer: Our required probability is 0.91.

Step-by-step explanation:

Since we have given that

n = 85

mean = 100 points

standard deviation = 11

We need to find the probability that the sample mean will not differ from the population mean by more than 2 points.

so, it becomes,

$P(100-2<\bar{x}<100+2)\\\\=P(98<\bar{x}<102)\\\\=P(\dfrac{98-100}{\dfrac{11}{\sqrt{85}}}<Z<\dfrac{102-100}{\dfrac{11}{\sqrt{85}}})\\\\=P(-1.68<Z<1.68)\\\\=P(z<1.68)-P(z<-1.68)\\\\=0.9535-0.0465\\\\=0.9070\\\\\approx 0.91$

Hence, our required probability is 0.91.