Exercise 12-10 asked you to assume that the actual weight of a certain candy bar

Question

Exercise 12-10 asked you to assume that the actual weight of a certain candy bar, whose advertised weight is 2.13 ounces, varies according to a normal distribution with mean mu=2.20 ounces and standard deviation= .04 ounces.

a. What is the probability that an individual candy bar weighs between 2.18 and 2.22 ounces?

b. What does the CLT say about how these sample means will vary from sample to sample?

The CLT says the sample means will be normally distributed with mean _____ ounces and standard deviation ______ ounces.

[Round your last answer to 4 decimal places.]

c. Use the Standard Normal Probabilities table or technology to calculate the probability that the sample mean weight of these five candy bars falls between 2.18 and 2.22 ounces.

Probability = _____

d. How do you expect this probability to change if the sample size were 40 instead of 5? Calculate the probability.

Probability= _____

Explanation / Answer

a. P(2.18<x<2.22)=P(2.18-2.20/0.04<z<2.22-2.20/0.04)=P(-0.5<z<0.5)=0.3829

b. As per CLT mean=2.20 and sd=0.04/sqrt(n)

c. P(2.18<xbar<2.22)=P(2.18-2.20/(0.04/sqrt(5))<z<2.22-2.20/(0.04/sqrt(5))=P(-1.12<z<1.12)=0.7373

d. Taking n=40,

P(-3.16<z<3.16)=0.9984

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