b). The intelligent Quotient (IQ) for a randomly selected female students at a c

Question

b).The intelligent Quotient (IQ) for a randomly selected female students at a certain University was measured by the Wechsler Intelligence Scale, and found to be a normal random variable X with a mean of µ=52.5 and a standard deviation of =10.

(i). What is the probability that a randomly chosen female has an IQ less than 37.5

(2 pts)

(ii). What is the probability that a randomly chosen female has an IQ less than 52.5 (2 pts)

(iii). What is the probability that a randomly chosen female has an IQ between 37.5 and 67.5 (2 pts)

(iv).An IQ is almost guaranteed with a probability of 0.950 to fall between what two values. (2 pts)

(v).The probability is only 0.0495 that an IQ will be greater than what score.

Explanation / Answer

. For the standard normal N(0,1)(z) or just N(z), z can be though of as the number of standard deviations from the mean, i.e. it is the distance from z = 0. So, to convert from a different µ and , we find out how far x is from the mean and divide by the standard deviation to get the number of standard deviations,
z = (x-µ)/

Given Z, we can use the standard normal tables (or our calculator, if equipped) to get the answer. If I just want the cumulative distribution table, i generally use
http://homes.cs.washington.edu/~jrl/norm...

1) Whats the probability that a randomly chosen female has an IQ less than 37.5?
Since our mean is 52.5 and our standard deviation is 10, we have
z = (37.5 - 52.5) / 10 = -1.5
So we want P(IQ < 37.5) = P(z < -1.5). Using the tables we find the answer is 0.0668 or 6.68%

2) Whats the probability that a randomly chosen female has an IQ less than 52.5?
z = (52.5 - 52.5) / 10 = 0
So we want P(IQ < 52.5) = P(z < 0). Using the tables we find the answer is 0.5 or 50%

3) Whats the probability that a randomly chosen female has an IQ between 37.5 and 67.5?
z1 = (37.5 - 52.5) / 10 = -1.5
We did this in (1) and the answer was 0.0668
z2 = (67.5 - 52.5) / 10 = 1.5
So the probability P(IQ < 67.5) = P(z2 < 1.5). Using the tables we find the answer is 0.9332.
So, since the probability that the IQ is less than 67.5 also includes that the IQ is less than 37.5, if we also want to have the IQ be greater than 37.5 we have to subtract that part of the probability. As a formula this says
P(a < z < b) = P(z < b) - P(z < a)
In our case this becomes
P( -1.5 < z < 1.5) = P(z < 1.5) - P(z < -1.5) = 0.9332 - 0.0668 = 0.8664 or 86.64%

(4) An IQ is almost guaranteed with a probability of 0.950 to fall between what two values?
We need to also assume that the question means the interval to be as much below the mean as above the mean or there is no unique answer. In addition we need a property of the normal distribution table which is
P(z < -a) = 1 - P( z < a)
For this case we need to work backwards. Looking at the last question and letting a = -b, we can use our formula to get, for the normal distribution,
P(-b < z < b) = P(z < b) - P(z < -b) = 2 P(z < b) - 1.
So, we want
2 P(z < b) - 1.= .95 or P(z < b) = 0.975
Looking in the table for 0.975 we see that b would be 1.96. Inverting the formula for z (working backwards), we have
x = z + µ = 10 * 1.96 + 52.5
So our answer is 72.1 on one side and
x = z + µ = 10 * (-1.96) + 52.5 = 32.9
on the other or
32.9 < IQ < 72.1

(5) The probability is only 0.0495 that an IQ will be greater than what score?
Again we need to find the for which P(z > b) = 0.0495. Since the probability the z will be greater than b and that z will be less than(or equal to) b is 1, we have
P( z > b) = 1 - P(z < b)
This is true for all probability functions (or a real variable). Thus
P(z > b ) = 1 - P(z < b) = .0495
or
P(z < b) = .9505
of b = 1.65. That can be converted to an IQ using our formula above.

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