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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.

Sagot :

lublana

Answer:

C.[tex]P(7.0039<x<7.8026)=0.1334[/tex]

D.[tex]P(7.0039<\bar{x}<7.8026)\approx 0.2942[/tex]

Step-by-step explanation:

We are given that

n=18

Mean, [tex]\mu=6.75[/tex]

Standard deviation, [tex]\sigma=2.28[/tex]

c.

[tex]P(7.0039<x<7.8026)=P(\frac{7.0039-6.75}{2.28}<\frac{x-\mu}{\sigma}<\frac{7.8026-6.75}{2.28})[/tex]

[tex]P(7.0039<x<7.8026)=P(0.11<Z<0.46)[/tex]

[tex]P(a<z<b)=P(z<b)-P(z<a)[/tex]

Using the formula

[tex]P(7.0039<x<7.8026)=P(Z<0.46)-P(Z<0.11)[/tex]

[tex]P(7.0039<x<7.8026)=0.67724-0.54380[/tex]

[tex]P(7.0039<x<7.8026)=0.1334[/tex]

D.[tex]P(7.0039<\bar{x}<7.8026)=P(\frac{7.0039-6.75}{2.28/\sqrt{18}}<\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}})<\frac{7.8026-6.75}{2.28/\sqrt{18}})[/tex]

[tex]P(7.0039<\bar{x}<7.8026)=P(0.47<Z<1.96)[/tex]

[tex]P(7.0039<\bar{x}<7.8026)=P(Z<1.96)-P(Z<0.47)[/tex]

[tex]P(7.0039<\bar{x}<7.8026)=0.97500-0.68082[/tex]

[tex]P(7.0039<\bar{x}<7.8026)=0.29418\approx 0.2942[/tex]

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