Question

Let x be a continuous random variable that follows a normal distribution with a mean of 185 and a standard deviation of 18. Find the value of x so that the area under the normal curve between μ and x is approximately 0.4812 and the value of x is greater than μ.

Answer 1

$$P(Z<\dfrac{x-\mu}{\sigma})-0.5=0.4812$$

$$P(Z<\dfrac{x-\mu}{\sigma})=0.9812$$

$$\dfrac{x-\mu}{\sigma}\approx2.079189$$

$$x\approx\mu+2.079189\sigma$$

Given $\mu=185, \sigma=18.$

$$x\approx185+2.079189(18)$$

$$x\approx222.425$$