# Thread: Basic Statistics Word Problems

1. ## Basic Statistics Word Problems

I have these problems due tomorrow - any help is greatly appreciated

1. Suppose that a brewery has a filling machine that fills 12 oz bottles of beer. It is known that the amount of beer poured by this filling machine follows a normal distribution with a mean 12.32 oz and a standard deviation of 0.04 oz. Find the probability that a bottle contains between 12.22 and 12.28 oz.

2. One year, pro sports players salaries averaged $1.5 million with a standard deviation of$0.6 million. Suppose a sample of 400 major league players was taken. Find the approximate probability that the average salary of the 400 players exceeded $1.1 million. 3. Assume that the salaries of elementary school teachers in the US are normally distributed with a mean of$32,000 and a standard deviation of $6,000. What is the cutoff salary for teachers in the bottom 10%? 4. The tread life of a particular brand of tire is a random variable best described by a normal distribution with mean of 60,000 miles and a standard deviation of 1,400 miles. What warranty should the company use if they want 96% of the tires to outlast the warranty? 2. Originally Posted by Guarana I have these problems due tomorrow - any help is greatly appreciated 1. Suppose that a brewery has a filling machine that fills 12 oz bottles of beer. It is known that the amount of beer poured by this filling machine follows a normal distribution with a mean 12.32 oz and a standard deviation of 0.04 oz. Find the probability that a bottle contains between 12.22 and 12.28 oz. Compute the z scores of 12.22 and 12,28 ounces, they are: z1 = (12.22-12.32)/0.04 = -2.5 and: z2 = (12.28-12.32)/0.04 = -1 Now look these up in a table of the cumulative standard normal distribution: P(z<z1) = P(z<-2.5) = 0.0062 P(z<z2) = P(z<-1) = 0.1587 So: P(z1<z<z2)= P(z<z2) - P(z<z1) = 0.1587 - 0.0062 = 0.1525 RonL 3. Originally Posted by Guarana 2. One year, pro sports players salaries averaged$1.5 million with a standard deviation of $0.6 million. Suppose a sample of 400 major league players was taken. Find the approximate probability that the average salary of the 400 players exceeded$1.1 million.
For a sample of size 400 the avaerage salary will have a mean m equal to the
population mean, and a standarn deviation equal to the population SD divided
by the square root of the sample size.

Therefore the avaerage of our sample is a random variable with mean $1.5 million, and SD$0.6/sqrt(400) ~= $0.03 million Thus the z-score corresponding to$1.1 million is:

z1 = (1.1 - 1.5)/0.03 = -13.3

The approximate value of P(z<z1) = P(z<-13.3) ~= 0, so P(z>z1) ~=1.

RonL