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A restaurant server believes the distribution of their tips has a model that is slightly skewed to the left, with a mean of $10.70 and a standard deviation of $6.70. They

usually wait on about 60 parties over a weekend of work.
a) Estimate the probability that they will earn at least $800 in tips.
b) How much do they earn on the best 10% of such weekends?

User Gad D Lord
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Answer:

a) To estimate the probability that the server will earn at least $800 in tips, we need to use the cumulative distribution function (CDF) of the normal distribution, since the mean and standard deviation of the tips are given. The CDF of the normal distribution with mean μ and standard deviation σ is given by F(x) = P(X <= x) = (1/2)(1 + erf((x-μ)/(σ * sqrt(2)))

We can use this formula to find the probability that the server will earn at least $800 in tips. We know that the mean of the tips is $10.70, and the standard deviation is $6.70. Therefore, we can calculate the CDF of the normal distribution with μ = $10.70 and σ = $6.70.

P(X >= 800) = 1 - P(X <= 800) = 1 - F(800)

b) To find out how much the server earns on the best 10% of such weekends, we need to find out what the 90th percentile of the distribution is. We can use the inverse cumulative distribution function (ICDF) of the normal distribution to find this. The ICDF of the normal distribution with mean μ and standard deviation σ is given by x = μ + σ * sqrt(2) * erf^-1(2F(x) - 1)

We know that the mean of the tips is $10.70, and the standard deviation is $6.70. Therefore, we can calculate the ICDF of the normal distribution with μ = $10.70 and σ = $6.70.

Then we can find the 90th percentile: x = $10.70 + $6.70 * sqrt(2) * erf^-1(2*0.1 - 1)

Please note that the above calculation required erf^-1(inverse error function) which is not a standard function on most calculators, you may need to use a software/programming language to find it.

User MalditoBarbudo
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