Question

If scores are normally distributed with the mean of 30 and standard deviation of 5. what percent score is.

Answer 1

(a)

$$P(X>30)=0.5$$

(b)

$$P(X>37)=1-P(Z\le\dfrac{37-30}{5})=0.0808$$

(c)

$$P(28<X<34)=P(Z<\dfrac{34-30}{5})$$

$$-P(Z\le\dfrac{28-30}{5})\approx0.4436$$