Answer:
(a) P(X > 95) = 63.1%
(b) P(X < 125) = 95.2%
(c) 523.25 people would have an IQ less than 110.
(d) 697.2 people would have an IQ greater than 140.
Step-by-step explanation:
We are given that IQ is normally distributed with a mean of 100 and a standard deviation of 15.
Let X = IQ level of a person
So, X ~ N()
The z score probability distribution is given by;
Z = ~ N(0,1)
where, = population mean
= population standard deviation
(a) Probability that this person has an IQ greater than 95 is given by = P(X > 95)
P(X > 95) = P( > ) = P(Z > -0.33) = P(Z < 0.33) = 0.631 or 63.1%
(b) Probability that this person has an IQ less than 125 is given by = P(X < 125)
P(X < 125) = P( < ) = P(Z < 1.67) = 0.9525 or 95.2%
(c) Now as we are given with the sample of 700 people. So, the z score probability distribution is now given as;
Z = ~ N(0,1)
where, = population mean
= population standard deviation
n = sample size = 700
So,<em> Probability of people that have an IQ less than 110 = P(X < 110) </em>
P(X < 110) = P( < ) = P(Z < 0.67) = 0.7475 or 74.75%
Now, people having an IQ less than 110 in a sample of 700 = 74.75% of 700
= = 523.25
Therefore, 523.25 people would have an IQ less than 110.
(d) Probability of people that have an IQ greater than 140 = P(X > 140)
P(X > 140) = P( > ) = P(Z > 2.67) = 0.9962 or 99.6%
Now, people having an IQ greater than 140 in a sample of 700 = 99.6% of 700
= = 697.2
Therefore, 697.2 people would have an IQ greater than 140.