IQ scores areknown to be normally distributed with mean 100 adn standard deviation 15?

Question

a certain class of 36 students is observed to have a mean iq of 106. if thesse 36 students were drawn at random form the general population, what would be the probability tht the mean iq score exceeds 106?

Answer 1

If you assume that the IQ of a person chosen at random in the population is a normal random variable with mean 100 and variance 15^2=225 then the mean IQ of 36 persons chosen at random is a normal random variable with mean 100 and variance 225/36=(15/6)^2=(2.5)^2

The probability you're looking for is the probability that
the mean is larger than 106
which equals the probability that
(the mean - 100)/(2.5) is larger than (106-100)(2.5)
but (the mean - 100)/(2.5) is a standard normal variable (mean=0, variance=1)
so you're looking for the probability that a standard normal variable is larger than
(106-100)/(2.5)=6/(2.5)=2.4

You'll find that a standard normal variable has a probability 0.99180 of being smaller than 2.4, hence a probability 1-0.99180=0.0082 of being larger.

The probability you're looking for is approx. 0.8%

Answer 2

The IQ test is a relative test that should have the mean scoring 100. If you have a mean that scores 106, then you probably need to change the scoring to equal 100. But then again, the IQ test is really meant to find retardation, not higher levels of intelligences. so if you have a group that scores moderately higher, probably it means nothing since the test wasn't designed to for this anyway... maybe. I've been drinking.. probably my IQ is a little low now.

Answer 3

Usually 32 is enough to ensure very little variability in a standard normal group. So with 36 (independent?) risks, standard dev of the group is 2.5. Z(6/2.5) = Z(2.4) = .9918 percent of being less than or equal to 106. So 0.89% of being above 106.

Answer 4

Convert 106 to a Z value and look up P(z
What you want is 1-P(z