A waiter believes the distribution of his tips has a model that is slightly skewed to the right, with a mean of $10.50 and a standard deviation of $5.20. 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) The probability that he will earn at least $600 is 0.0212.

(b) The amount of tip the waiter earns on the best 1% of such weekends is $610.67.

Step-by-step explanation:

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,

The random variable X can be defined as the tips he receiver per order.

The average tip received by the waiter is, μ = $10.50.

The standard deviation of the tip received by the waiter is, σ = $5.20.

The waiter usually waits on about n = 50 parties over a weekend of work.

So, the distribution of the total tip earned by the waiter is:

.

(a)

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

*Use a z-table for the probability.

Thus, the probability that he will earn at least $600 is 0.0212.

(b)

Let be a be the amount of tip the waiter earns on the best 1% of such weekends.

That is,

P (∑X > a) = 0.01

⇒ P (∑X < a) = 0.99

⇒ P (Z < z) = 0.99

The value of z for the above probability is:

z = 2.33.

Compute the value of a as follows:

Thus, the amount of tip the waiter earns on the best 1% of such weekends is $610.67.