Question

X is a normally distributed random variable with mean 95 and standard deviation 3. What is the probability that X is between 89 and 98? Write your answer as a decimal rounded to the nearest thousandth.

Answer 1

Answer: 0.819

Explanation:

Given: X is a normally distributed random variable with mean 95 and standard deviation 3.

Then, the probability that X is between 89 and 98 will be:

`P(89<X<95)=P((89-95)/(3)<(X-mean)/(standard\ deviation)<(98-95)/(3))\\\\=P(-2<Z<1)\ \ \ [Z=(X-mean)/(standard\ deviation)]\\\\ =P(Z<1)-(1-P(Z<2))\\\\= 0.8413- (1-0.9772)\ \ \ [\text{By p-value table}]\\\\= 0.8185\approx0.819`

Hence, the required probability = 0.819