Question

1) Bringing all of section 6.2 together...

A company knows that replacement times for the DVD players it produces are Normally distributed with a mean of 4.5 years and a standard deviation of 1.1 years.

Find the proportion of a randomly selected DVD players that will have a replacement time less than 2 years?
P(X < 2 years) = _________________ (Enter your answer accurate to 4 decimal places.)

If the company wants to provide a warranty so that only 3.1% of the DVD players will be replaced before the warranty expires, what is the time length of the warranty?
warranty = ___________years (Enter your answer as a number accurate to 1 decimal place.)

2) For a standard normal distribution, find:

P(z > c) = 0.0286

Find c. In other words, find the z-score that has 0.0286 to its right.

_____________

3)Assume that the readings at freezing on a batch of thermometers are Normally distributed with mean 0°C and standard deviation 1.00°C.

Find P₈₀, the 80-percentile of the distribution of temperature readings. This is the temperature reading separating the bottom 80% from the top 20%.

________________°C Round to 3 decimal places.

4)For a standard normal distribution, given:

P(z < c) = 0.8787

Find c. In other words, find the z-score that has 0.8787 to its left.

____________

Answer 1

Solution - l=4.5 D 82=Xuly a) P(x22) = P(2.2- p(222-4.5) lol Plz Z2-2.27) 0.0116 zo-8 = -1-87 b) T-score for lowest 3.1.1. of score Z = X-M -1.87 = X-4.5 lol - ) x = 2.4 2) P/2> c) = 0.0206 1.90 с 2 3) P12(70) = 0.80 0.842 =) Zo= 4) P (220) = 0.8787 ca 1.17