Answer to Question #328613 in Statistics and Probability for Sevi

Question

1. A certain research center requires an IQ score above 131.5.

Assume IQ scores are normally distributed with mean of 100 and standard

deviation of 15. Nine candidates took IQ tests. Find the probability that the IQ

score of a randomly chosen individual is at least 131.5?

2. A Company which produces cigarettes claims that it has now reduced the

amount of nicotine. The Supporting evidence consists of a random sample of 40

cigarettes and gives mean amount of nicotine 0.941 g and a standard deviation of

0.313g. What is the probability that mean nicotine amount is greater than 0.882 g.

Answer 1

$1:\\P\left( X\geqslant 131.5 \right) =P\left( \frac{X-100}{15}\geqslant \frac{131.5-100}{15} \right) =P\left( Z\geqslant 2.1 \right) =\\=\varPhi \left( -2.1 \right) =0.0179\\2:\\P\left( \bar{x}>0.882 \right) =P\left( \sqrt{n}\frac{\bar{x}-\mu}{\sigma}>\sqrt{n}\frac{0.882-\mu}{\sigma} \right) =\\=P\left( Z>\sqrt{40}\frac{0.882-0.941}{0.313} \right) =P\left( Z>-1.19217 \right) =\\=\varPhi \left( 1.19217 \right) =0.8834$