Answer to Question #139055 in Statistics and Probability for Mla

Question

the 2019 bar exams only 2103 of the 7685 bar passed. The passing a weighted average of 74, which is higher than the mean weighted average of all the bar examinees. Find this mean given that the weighted average is normally distributed with a standard deviation of 16.62. Round off the final answer to the nearest whole number

Answer 1

Let $X=$ the weighted average: $X\sim N(\mu, \sigma^2).$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0,1)$

$$P(X>74)=1-P(X\leq74)=$$

$$=1-P(Z\leq\dfrac{74-\mu}{16.62})=\dfrac{2103}{7685}$$

$$P(Z\leq\dfrac{74-\mu}{16.62})=\dfrac{5582}{7685}\approx0.72635$$

$$\dfrac{74-\mu}{16.62}\approx0.601811$$

$$\mu\approx64$$