Question

NEED HELP

The average amount of money spent for lunch per person in the college cafeteria is $6.75 and the standard deviation is $2.28. Suppose that 18 randomly selected lunch patrons are observed. Assume the distribution of money spent is normal, and round all answers to 4 decimal places where possible.


C. For a single randomly selected lunch patron, find the probability that this

patron's lunch cost is between $7.0039 and $7.8026.

D. For the group of 18 patrons, find the probability that the average lunch cost is between $7.0039 and $7.8026.

Answer 1

C.
`P(7.0039<x<7.8026)=0.1334`

D.
`P(7.0039<\bar{x}<7.8026)\approx 0.2942`

Explanation:

We are given that

n=18

Mean,
`\mu=6.75`

Standard deviation,
`\sigma=2.28`

c.

`P(7.0039<x<7.8026)=P((7.0039-6.75)/(2.28)<(x-\mu)/(\sigma)<(7.8026-6.75)/(2.28))`

`P(7.0039<x<7.8026)=P(0.11<Z<0.46)`

`P(a<z<b)=P(z<b)-P(z<a)`

Using the formula

`P(7.0039<x<7.8026)=P(Z<0.46)-P(Z<0.11)`

`P(7.0039<x<7.8026)=0.67724-0.54380`

`P(7.0039<x<7.8026)=0.1334`

D.
`P(7.0039<\bar{x}<7.8026)=P((7.0039-6.75)/(2.28/√(18))<(x-\mu)/((\sigma)/(√(n))))<(7.8026-6.75)/(2.28/√(18)))`

`P(7.0039<\bar{x}<7.8026)=P(0.47<Z<1.96)`

`P(7.0039<\bar{x}<7.8026)=P(Z<1.96)-P(Z<0.47)`

`P(7.0039<\bar{x}<7.8026)=0.97500-0.68082`

`P(7.0039<\bar{x}<7.8026)=0.29418\approx 0.2942`