A waitress believes the distribution of her tips has a model that is slightly skewed to the left, with a mean of $9.40 and a standard deviat

Question

A waitress believes the distribution of her tips has a model that is slightly skewed to the left, with a mean of $9.40 and a standard deviation of $6.10. She usually waits on about 60 parties over a weekend of work.

a) Estimate the probability that she will earn at least $600

P(tips from 60 parties > $600)

b) How much does she earn on the best 5% of such weekends?

The total amount that she earns on the best 5% of such weekends is at least $__

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Anna 1 month 2021-10-18T13:31:35+00:00 1 Answer 0 views 0

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    2021-10-18T13:32:42+00:00

    Answer:

    a. P(tips from 60 parties > $600)=0.461

    b. The total amount that she earns on the best 5% of such weekends is at least $19.43.

    Step-by-step explanation:

    a. To earn $600 in 60 parties means $10 per party in average.

    If we assume a normal distribution of tips, we can calculate the z-value and its probability for this situation:

    z=\frac{x-\mu}{\sigma}=\frac{10.00-9.40}{6.10}=\frac{0.60}{6.10}=   0.098\\\\P(x>10)=P(z>0.098)=0.461

    There is a probability of 46% that she earns at least $600 over a weekend of work.

    b. The best 5% of the weekends corresponds to:

    P(x>x_1)=0.05

    This probability (5%) corresponds to a z-value of z=1.6449.

    In tips, this value represents:

    x=\mu+z*\sigma=9.40+1.6449*6.10=9.40+10.03=19.43

    The total amount that she earns on the best 5% of such weekends is at least $19.43.

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