# The charge life of a certain lithium ion battery for camcorders is normally distributed, with mean u=90 minutes and standard deviation o=35 minutes. a. what is the probability that a randomly...

The charge life of a certain lithium ion battery for camcorders is normally distributed, with mean u=90 minutes and standard deviation o=35 minutes.

a. what is the probability that a randomly selected battery of this type lasts more than 100 minutes on a single charge? Is this result unusual?

b. Describe the sampling distribution of sample mean (_ over x), the sample mean charge life for a random sample of 10 such batteries.

c. What is the probability that a random sample of 10 such batteries lasts more than 100 minutes? Is this result unusual?

d. What is the probability that a random sample of 25 such batteries has a mean charge of more than 100 minutes?

e. Explain what causes that probabilities in parts (c) and (d) to be different.

### 1 Answer | Add Yours

Given `mu=90,sigma=35` :

(a) Find `P(x>100)`

Convert the datum to a standard score: `z=(100-90)/35=.29`

Then `P(x>100)=P(z>.29)` . From the standard normal table we find the area to the right of z=.29 to be .3859

**So `P(x>100)~~.3859` . It happens about 39% ofthe time, so not too unusual.**

(b) For a sample of `n=10` the sampling distribution has `bar(x)=90,s=35/sqrt(10)~~11.07`

(c) For a sample of size 10 find `P(bar(x)>100)`

Again we convert to a standardized score: `z=(100-90)/(35/sqrt(10))~~.9035`

(Note the use of the standard error)

`P(bar(x)>100)=P(z>.9035)` .The area under the standard normal curve to the right of z=.90 is .1841 (from a standard normal table)

Thus `P(bar(x)>100)~~.1841`

(d) For n=25 find `P(bar(x)>100)`

`z=(100-90)/(35/sqrt(25))~~1.43`

`P(bar(x)>100)=P(z>1.43)=.0764`

(e) Since the sample in (d) is larger, the standard error is smaller. As the sample size increases, the means of the samples get closer to the true population mean (the error decreases).

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