Question

Given the population of 5000 scores with a mean of 86 and standard deviation of 10. How many scores

Answer 1

a)

$$P(86<X<96)=P(Z<\dfrac{96-86}{10})-P(Z\leq\dfrac{86-86}{10})$$

$$=P(Z<1)-0.5=0.341345$$

$$0.341345(5000)=1706.7\approx1707$$

b)

$$P(96<X<106)=P(Z<\dfrac{106-86}{10})-P(Z\leq\dfrac{96-86}{10})$$

$$=P(Z<2)-P(Z\leq 1)=0.977250-0.841345$$

$$=0.135905$$

$$0.135905(5000)=679.5\approx680$$

c)

$$0.56(5000)=2800$$

2800 scores are below 56%

$$P(X<56)=P(Z<\dfrac{56-86}{10})=P(Z<-3)$$

$$=0.001350$$

$$0.001350(5000)=6.75\approx7$$

7 scores are below 56.