Iq scores are normally distributed with a mean of 100 and a standard deviation of 15. to the nearest integer values how high must your in be to fall within the top 5% of scores

124.75

Step-by-step explanation:

Lets X be aleatory variable for iq scores. Here, wee need a value α such that, the probability that X is greater than α is higher than 0.95. This is:

P(X > α)=0.95

As X distributes normal, if we subtract the mean and divide y its std. dev. it will distribute normal with mean 0 and sts. dev. 1, which permits us to use the normal distribution table to solve it. Doing so:

P([X - 100]/15 > [α-100]/15)=0.95

Using the normal table, we need to look for the value that gives us a probability on 0.95, an it it 1.65.

So, 1.65 must be equal to [α-100]/15 :

1.65 = [α-100]/15

Multiplying both sides by 15:

24.75 = α-100

Summing 100 in both sides:

124.75 = α

So, for scores equals or greater to 124.75 you will be in the top 5% scores.

homozygous. one is homozygous, while two is heterozygous.

the answer is c.

step-by-step explanation: