Respuesta :
Answer:
(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,
[tex]\mu_{S}=n\mu\\[/tex]
And the standard deviation of the distribution of the sum of values of X is given by,
[tex]\sigma_{S}=\sqrt{n}\sigma[/tex]
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:
[tex]\mu_{S}=50\times 10.60=530\\\\\sigma_{S}=\sqrt{50}\times 6.60=46.67[/tex]
(a)
Compute the probability that he will earn at least $600 as follows:
[tex]P(S\geq 600)=P(\frac{S-\mu_{S}}{\sigma_{S}}\geq \frac{600-530}{46.67})\\\\=P(Z>1.50)\\\\=1-P(Z<1.50)\\\\=1-0.93319\\\\=0.06681\\\\\approx 0.0668[/tex]
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:
[tex]z=\frac{S-\mu_{S}}{\sigma_{S}}\\\\2.33=\frac{x-530}{46.67}\\\\x=530+(2.33\times 46.67)\\\\x=638.7411\\\\x\approx 638.74[/tex]
Thus, on the best 1% of such weekends the waiter earned $638.74.
