Question

A distribution of scores is approximately normal with a mean of 78 and a standard deviation of 8.6. Write an equation can be used to find the score x above which 33 percent of the scores fall?

Answer 1

Answer:

$\displaystyle\frac{x - 78}{8.6} = -0.440$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 78

Standard Deviation, σ = 8.6

We are given that the distribution is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.33

$P( X > x) = P( z > \displaystyle\frac{x - 78}{8.6})=0.33$  

$= 1 -P( z \leq \displaystyle\frac{x - 78}{8.6})=0.33$  

$=P( z \leq \displaystyle\frac{x - 78}{8.6})=0.67$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x - 78}{8.6} = -0.440\\\\x = 74.22$  

which is the required equation.

Thus, values greater than equal to 74.22 above which 33% of the score fall.