Answer:
(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 <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sum of values of <em>X</em>, i.e ∑<em>X</em>, will be approximately normally distributed.
Then, the mean of the distribution of the sum of values of <em>X</em> is given by,
And the standard deviation of the distribution of the sum of values of <em>X</em> is given by,
The random variable <em>X</em> can be defined as the tips he receiver per order.
The average tip received by the waiter is, <em>μ</em> = $10.50.
The standard deviation of the tip received by the waiter is, <em>σ</em> = $5.20.
The waiter usually waits on about <em>n</em> = 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 <em>z-</em>table for the probability.
Thus, the probability that he will earn at least $600 is 0.0212.
(b)
Let be <em>a</em> 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 <em>z</em> for the above probability is:
<em>z</em> = 2.33.
Compute the value of <em>a</em> as follows:
Thus, the amount of tip the waiter earns on the best 1% of such weekends is $610.67.