Question

A large part of the luggage market is made up of overnight bags. These bags vary by weight, exterior appearance, material, and size. Suppose the volume of overnight bags is normally distributed with mean mu = 1750 cubic inches and standard deviation sigma = 250 cubic inches. A random sample of 15 overnight bags is selected, and the volume of each is found.

a. Find the distribution of X bar.
b. What is the probability that the sample mean volume is more than 1800 cubic inches?
c. What is the probability that the sample mean volume is within 100 cubic inches of 1750?
d. Find a symmetric interval about 1750 such that 95% of all values of the sample mean volume lie in this interval.

Answer 1

a)
`\bar X \sim N(\mu=1750, (\sigma)/(√(n))=(250)/(√(15))=64.550)`

b)
`P(\bar X >1800)=P(Z>(1800-1750)/((250)/(√(15)))=0.775)`

Using the complement rule and the normal standard table or excel we have:

`P(Z>0.775)=1-P(Z<0.775)= 1-0.781=0.219`

c)
`P(1650<\bar X <1850)=P((1650-1750)/((250)/(√(15)))<Z<(1850-1750)/((250)/(√(15))))= P(-1.549<Z<1.549)`

We can find this probability with the following operation:

`P(-1.549< Z< 1.549) = P(Z<1.549)-P(Z<-1.549) = 0.939-0.061= 0.879`

d)
`1750-1.96(250)/(√(15))=1623.48`

`1750+1.96(250)/(√(15))=1876.517`

So on this case the 95% confidence interval would be given by (1623.48;1876.517)

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the volume of overnight bags of a population, and for this case we know the distribution for X is given by:

`X \sim N(1750,250)`

Where
`\mu=1750` and
`\sigma=250`

Since the distribution for X is normal the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu=1750, (\sigma)/(√(n))=(250)/(√(15))=64.550)`

Part b

For this case we want this probability:

`P(\bar X >1800)`

We can use the z score given by:

`z= (\bar X -\mu)/((\sigma)/(√(n)))`

Using this formula we got:

`P(\bar X >1800)=P(Z>(1800-1750)/((250)/(√(15)))=0.775)`

Using the complement rule and the normal standard table or excel we have:

`P(Z>0.775)=1-P(Z<0.775)= 1-0.781=0.219`

Part c

For this case we want this probability:

`P(1650<\bar X <1850)`

Using the z score formula we got:

`P(1650<\bar X <1850)=P((1650-1750)/((250)/(√(15)))<Z<(1850-1750)/((250)/(√(15))))= P(-1.549<Z<1.549)`

We can find this probability with the following operation:

`P(-1.549< Z< 1.549) = P(Z<1.549)-P(Z<-1.549) = 0.939-0.061= 0.879`

Part d

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The confidence interval for the mean is given by the following formula:

`\bar X \pm z_(\alpha/2)(\sigma)/(√(n))` (1)

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that
`z_(\alpha/2)=1.96`

Now we have everything in order to replace into formula (1):

`1750-1.96(250)/(√(15))=1623.48`

`1750+1.96(250)/(√(15))=1876.517`

So on this case the 95% confidence interval would be given by (1623.48;1876.517)