Suppose that scores on a particular test are normally distributed with a mean of 140 and a standard deviation of 18. What is the minimum score needed to be in the top 5% of the scores on the test? Carry your intermediate computations to at least four decimal places, and round your answer to one decimal place.

153.16

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = ???

μ is the population mean = 140

σ is the population standard deviation = 8

Top 5% means the score is in the 95th percentile

The z score of 95th percentile = 1.645

1.645 = x - 140/8

Cross Multiply

1.645 × 8 = x - 140

13.16 = x - 140

x = 13.16 + 140

x = 153.16

Therefore, the minimum score needed to be in the top 5% of the scores on the test is 153.16

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6)   ·x^(1/6) = x^(4/6)   since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)