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prohojiy [21]
3 years ago
13

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?
Mathematics
1 answer:
Vitek1552 [10]3 years ago
7 0

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

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

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