Answer to Question #305020 in Statistics and Probability for Sunitha

Question

In a normal distribution 41% of items are under 55 and 11% are over 66. Find the mean and variance of the distribution.

Answer 1

$$P(Z<\dfrac{55-\mu}{\sigma})=0.41$$

$$\dfrac{55-\mu}{\sigma}\approx-0.227545$$

$$P(Z>\dfrac{66-\mu}{\sigma})=0.11$$

$$\dfrac{66-\mu}{\sigma}\approx1.226528$$

$$\mu=66-1.226528\sigma$$

$$66-1.226528\sigma=55+0.227545\sigma$$

$$\sigma=7.565$$

$$\mu=56.72$$