Question

The mean IQ score for 2000 students is 100​, with a standard deviation of 10. Assuming the scores have a normal​ curve, answer the following.
a. How many have an IQ score between 90 and 110​?
b. How many have an IQ score between 80 and 120​?
c. How many have an IQ score over 130​?

Answer 1

Given $\mu=100, \sigma=10$

a.

$$P(90<X<110)=P(X<110)-P(X\leq90)$$

$$=P(Z<\dfrac{100-100}{10})-P(Z\leq\dfrac{90-100}{10})$$

$$=P(Z<0)-P(Z\leq-1)\approx0.5-0.158655$$

$$=0.341345$$

$$0.341345(2000)=683$$

683 students

b.

$$P(80<X<120)=P(X<120)-P(X\leq80)$$

$$=P(Z<\dfrac{120-100}{10})-P(Z\leq\dfrac{80-100}{10})$$

$$=P(Z<2)-P(Z\leq-2)\approx0.977250-0.022750$$

$$=0.9545$$

$$0.9545(2000)=1909$$

By the 68-95-99.7 rule

$$P(\mu-2\sigma\leq X\leq\mu+2\sigma)\approx0.9545$$

$$0.9545(2000)=1909$$

1909 students

c.

$$P(X>130)=1-P(X\leq130)$$

$$=1-P(Z\leq\dfrac{130-100}{10})=1-P(Z\leq 3)$$

$$\approx1-0.998650=0.00135$$

$$0.00135(2000)=3$$

3 students