Respuesta :
Answer:
0.0015 = 0.15% probability that if the motel books 150 guests, not enough seats will be available.
Step-by-step explanation:
Binomial probability distribution
Probability of exactly x successes on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex], if [tex]np \geq 10[/tex] and [tex]n(1-p) \geq 10[/tex].
150 guests booked:
This means that [tex]n = 150[/tex]
85% of booked guests show up for their room.
This means that [tex]p = 0.85[/tex]
Is the normal approximation suitable:
[tex]np = 150(0.85) = 127.5[/tex]
[tex]n(1-p) = 150(0.15) = 22.5[/tex]
Both greater than 10, so yes.
Mean and standard deviation:
[tex]\mu = E(X) = np = 150*0.85 = 127.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{150*0.85*0.15} = 4.3732[/tex]
Find the probability that if the motel books 150 guests, not enough seats will be available.
More than 140 show up, which, using continuity correction, is [tex]P(X > 140 + 0.5) = P(X > 140.5)[/tex], which is 1 subtracted by the p-value of Z when X = 140.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{140.5 - 127.5}{4.3732}[/tex]
[tex]Z = 2.97[/tex]
[tex]Z = 2.97[/tex] has a p-value of 0.9985.
1 - 0.9985 = 0.0015.
0.0015 = 0.15% probability that if the motel books 150 guests, not enough seats will be available.
