Question

A consumer survey indicates that the average household spendsμ= $185on groceries each week. The distribution of spending amounts is approximately normal with a standard deviation of σ= $25. What proportion of the population spends more than $200 per week on grocaries?

Answer 1

Answer:

$P(X>200)=P(\frac{X-\mu}{\sigma}>\frac{200-\mu}{\sigma})=P(Z>\frac{200-185}{25})=P(Z>0.6)$

And we can find this probability using the complement rule and with the normal standard table or excel:

$P(Z>0.6)=1-P(z<0.6)=1-0.7257=0.2743$

Step-by-step 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".  

Solution to the problem

Let X the random variable that represent the average household spent of a population, and for this case we know the distribution for X is given by:

$X \sim N(185,25)$  

Where $\mu=185$ and $\sigma=25$

We are interested on this probability

$P(X>200)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>200)=P(\frac{X-\mu}{\sigma}>\frac{200-\mu}{\sigma})=P(Z>\frac{200-185}{25})=P(Z>0.6)$

And we can find this probability using the complement rule and with the normal standard table or excel:

$P(Z>0.6)=1-P(z<0.6)=1-0.7257=0.2743$