Question

The Insurance Institute reports that the mean amount of life insurance per household in the US is $110,000. This follows a normal distribution with a standard deviation of $40,000.

Answer 1

a)
`\sigma_(\bar X) = (\sigma)/(√(n))= (40000)/(√(50))= 5656.85`

b) Since the distribution for X is normal then we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

c)
`P( \bar X >112000) = P(Z>(112000-110000)/((40000)/(√(50)))) = P(Z>0.354)`

And we can use the complement rule and we got:

`P(Z>0.354) = 1-P(Z<0.354) = 1-0.600 = 0.400`

d)
`P( \bar X >100000) = P(Z>(100000-110000)/((40000)/(√(50)))) = P(Z>-1.768)`

And we can use the complement rule and we got:

`P(Z>-1.768) = 1-P(Z<-1.768) = 1-0.0385 = 0.962`

e)
`P(100000< \bar X <112000) = P(<(100000-110000)/((40000)/(√(50)))<Z<(112000-110000)/((40000)/(√(50)))) = P(-1.768<Z<0.364)`

And we can use the complement rule and we got:

`P(-1.768<Z<0.364) = P(Z<0.364)-P(Z<-1.768) = 0.642-0.0385 = 0.604`

Explanation:

a. If we select a random sample of 50 households, what is the standard error of the mean?

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".

Let X the random variable that represent the amount of life insurance of a population, and for this case we know the distribution for X is given by:

`X \sim N(110000,40000)`

Where
`\mu=110000` and
`\sigma=40000`

If we select a sample size of n =35 the standard error is given by:

`\sigma_(\bar X) = (\sigma)/(√(n))= (40000)/(√(50))= 5656.85`

b. What is the expected shape of the distribution of the sample mean?

Since the distribution for X is normal then we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

c. What is the likelihood of selecting a sample with a mean of at least $112,000?

For this case we want this probability:

`P(X > 112000)`

And we can use the z score given by:

`z= (\bar X  -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`P( \bar X >112000) = P(Z>(112000-110000)/((40000)/(√(50)))) = P(Z>0.354)`

And we can use the complement rule and we got:

`P(Z>0.354) = 1-P(Z<0.354) = 1-0.600 = 0.400`

d. What is the likelihood of selecting a sample with a mean of more than $100,000?

For this case we want this probability:

`P(X > 100000)`

And we can use the z score given by:

`z= (\bar X  -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`P( \bar X >100000) = P(Z>(100000-110000)/((40000)/(√(50)))) = P(Z>-1.768)`

And we can use the complement rule and we got:

`P(Z>-1.768) = 1-P(Z<-1.768) = 1-0.0385 = 0.962`

e. Find the likelihood of selecting a sample with a mean of more than $100,000 but less than $112,000

For this case we want this probability:

`P(100000<X < 112000)`

And we can use the z score given by:

`z= (\bar X  -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`P(100000< \bar X <112000) = P(<(100000-110000)/((40000)/(√(50)))<Z<(112000-110000)/((40000)/(√(50)))) = P(-1.768<Z<0.364)`

And we can use the complement rule and we got:

`P(-1.768<Z<0.364) = P(Z<0.364)-P(Z<-1.768) = 0.642-0.0385 = 0.604`