"A waiter believes the distribution of his tips has a model that is slightly skewed to the left​, with a mean of ​$10.60 and a standard deviation of ​$6.60. He usually waits on about 50 parties over a weekend of work. ​a) Estimate the probability that he will earn at least ​$600. ​b) How much does he earn on the best 1​% of such​ weekends?"

(a) 0.0668

(b) $638.74

Step-by-step explanation:

Let X denote the tips earned by a waiter.

It is provided that X follows a left-skewed distribution with mean, μ = $10.60 and standard deviation, σ = $6.60.

It is also provided that, the waiter usually waits on about n = 50 parties over a weekend of work.

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

And the standard deviation of the distribution of the sum of values of X is given by,  

As the sample size is large, i.e. n = 50 > 30, the Central Limit Theorem can be used to approximate the sampling distribution of total tips by the normal distribution.

The mean and standard deviation are:

(a)

Compute the probability that he will earn at least $600 as follows:

Thus, the probability that he will earn at least $600 is 0.0668.

(b)

Let x represents his earnings on the best 1%  of such weekends.

That is, P (X < x) = 0.99.

⇒ P (Z < z) = 0.99

The corresponding z-score is, 2.33.

Compute the value of x as follows:

Thus, on the best 1%  of such weekends the waiter earned $638.74.

step-by-step explanation:

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