Question

Suppose that the amount of a weekly grocery bill of all households in a metropolitan area is distributed with a mean of $140. If $35 is the standard deviation of amount of grocery bill of a random sample of 49 households selected from this metropolitan area, find the probability that average amount of grocery bill of such a sample will be(a) more than $145,(b) less than $140,(c) between $132 and $148.

Answer 1

a)
`z = (145-140)/((35)/(√(49)))=1`

We can use the complement rule and the normal standard table or excel and we got:

`P(Z>1) = 1-P(Z<1) = 1-0.841= 0.159`

b)
`z = (140-140)/((35)/(√(49)))=0`

We can use the normal standard table or excel and we got:

`P(Z<0) = 0.5`

c)
`z = (132-140)/((35)/(√(49)))=-1.6`

`z = (148-140)/((35)/(√(49)))=`

We can use the normal standard table or excel and we got:

`P(-1.6<Z<1.6) =P(Z<1.6)-P(Z<-1.6)=0.945-0.055= 0.890`

Explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let's assume that the random variable X represent the amount of weekly grocery bill of all households in a metropolitan area. We have the following info given:

`\mu = 140 , \sigma = 35`

From the central limit theorem we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

Part a

We want to calculate this probability:

`P(\bar X>145)`

And we can use the z score given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And if we use the value of 145 we got:

`z = (145-140)/((35)/(√(49)))=1`

We can use the complement rule and the normal standard table or excel and we got:

`P(Z>1) = 1-P(Z<1) = 1-0.841= 0.159`

Part b

`z = (140-140)/((35)/(√(49)))=0`

We can use the normal standard table or excel and we got:

`P(Z<0) = 0.5`

Part c

`z = (132-140)/((35)/(√(49)))=-1.6`

`z = (148-140)/((35)/(√(49)))=`

We can use the normal standard table or excel and we got:

`P(-1.6<Z<1.6) =P(Z<1.6)-P(Z<-1.6)=0.945-0.055= 0.890`