Question

A consumer survey indicates that the average household spends μ =$155 on groceries each week. The distribution of spending amounts is approximately normal with a standard deviation σ =$25. Based on this distribution,
What proportion of the population spends more than $175 per week on groceries?
How much money do you need to spend on groceries each week to be in the top 20% of the distribution?
How much does your family spend per week on groceries, what is your family�s percentile?

Answer 1

Answer:

a) 21.2%

b)  $176.05 or more

c) 15.87%    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = $155

Standard Deviation, σ = $25

We are given that the distribution of spending amounts is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(spends more than $175 per week on groceries)

P(x > 175)

$P( x > 175) = P( z > \displaystyle\frac{175 - 155}{25}) = P(z > 0.8)$

$= 1 - P(z \leq 0.8)$

Calculation the value from standard normal z table, we have,  

$P(x > 175) = 1 - 0.788 = 0.212 = 21.2\%$

b) P(X > x) = 0.2

We have to find the value of x such that the probability is 0.2

P(X > x)  

$P( X > x) = P( z > \displaystyle\frac{x - 155}{25})=0.2$  

$= 1 -P( z \leq \displaystyle\frac{x - 155}{25})=0.2$  

$=P( z \leq \displaystyle\frac{x - 155}{25})=0.8$  

Calculation the value from standard normal z table, we have,  

$P(z < 0.842) = 0.8$

$\displaystyle\frac{x - 155}{25} = 0.842\\\\x = 176.05$  

A consumer has to spend approximately $176.05 or greater to be in the top 20% of the distribution.

c) My family spends on average $130 dollars on groceries.

P(less than $130)

$P( x < 130) = P( z < \displaystyle\frac{130 - 155}{25}) = P(z < -1)$

Calculation the value from standard normal z table, we have,  

$P(x < 130) = 0.1587 = 15.87\%$

Thus, my family percentile is 15.87%