t A consumer survey indicates that the average household spends µ = $185 on groc

Question

t

A consumer survey indicates that the average household spends µ = $185 on groceries each week. The distribution of spending amounts is approximately normal with a standard deviation of = $25. Based on this distribution,

a . What proportion of the population spends more than $200 per week on groceries?

b . What is the probability of randomly selectinga family that spends less than $150 per week on groceries?

c . How much money do you need to spend on groceries each week to be in the top 20% of the distribution?

Please explain the steps.

Explanation / Answer

P(x>200)=0.2763

Explanation:

Step 1: Sketch the curve.

The probability that X>200 is equal to the blue area under the curve.

Step 2:

Since =185 and =25 we have:

P ( X>200 )=P ( X>200185 )

=P ( X)/>(200185)/25)

Since Z=(x)/ and( 200185)/25=0.6 we have:

P ( X>200 )=P ( Z>0.6 )

Step 3: Use the standard normal table to conclude that:

P (Z>0.6)=0.2743

B) P (X<150)=0.0808

Step 2:

Since =185 and =25 we have:

P ( X<150 )=P ( X<150185 )=P (X)/<(150185)/25)

Since (x)/=Z and (150185)/25=1.4 we have:

P (X<150)=P (Z<1.4)

Step 3: Use the standard normal table to conclude that:

P (Z<1.4)=0.0808

C)

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.