Respuesta :
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 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,
[tex]\mu_{X}=n\mu[/tex]
And the standard deviation of the distribution of the sum of values of X is given by,
[tex]\sigma_{X}=\sqrt{n}\sigma[/tex]
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:
[tex]\sum X\sim N(\$525,\ \$36.77)[/tex].
(a)
Compute the probability that he will earn at least $600 as follows:
[tex]P(\sum X\geq 600)=P(\sum X>600-0.5)[/tex]
[tex]=P(\sum X>599.5)\\[/tex]
[tex]=P(\frac{\sum X-\mu_{X}}{\sigma_{X}}>\frac{599.5-525}{36.77})[/tex]
[tex]=P(Z>2.03)\\=1-(Z<2.03)\\=1-0.97882\\=0.02118\\\approx0.0212[/tex]
*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:
[tex]z=\frac{a-\mu_{X}}{\sigma_{X}}[/tex]
[tex]2.33=\frac{a-525}{36.77}[/tex]
[tex]a=525+(2.33\times 36.77}\\a=610.6741\\a\approx610.67[/tex]
Thus, the amount of tip the waiter earns on the best 1% of such weekends is $610.67.
