Question

(Normal distribution: Finding a raw score)

Suppose that scores on a particular test are normally distributed with a mean of 110 and a standard deviation of 19. What is the minimum score needed to be in the top

10% of the scores on the test? Carry your intermediate computations to at least four decimal places, and round your answer to one decimal place.

Answer 1

µ = 110

sd = 19

$P(X>x)=0.1$

$or,P(\frac{X-\mu}{\sigma}>\frac{x-\mu}{\sigma})=0.1$

$or,P(Z>\frac{x-110}{19})=0.1$

$or,1-P(Z<\frac{x-110}{19})=0.1$

$or,P(Z<\frac{x-110}{19})=0.9$

$or,\frac{x-110}{19}=1.28155$

or, x = 134.3 (ans)