Respuesta :
Answer:
1.19% probability that fewer than 101 commuters drive to work alone
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses 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 samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, 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].
In this problem, we have that:
[tex]p = 0.75, n = 150[/tex]
[tex]\mu = E(X) = 150*0.75 = 112.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{150*0.75*0.25} = 5.3[/tex]
(a) What is the probability that fewer than 101 commuters drive to work alone
Using continuity corretion, this is [tex]P(X < 101-0.5) = P(X < 100.5)[/tex], which is the pvalue of Z when X = 100.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{100.5 - 112.5}{5.3}[/tex]
[tex]Z = -2.26[/tex]
[tex]Z = -2.26[/tex] has a pvalue of 0.0119
1.19% probability that fewer than 101 commuters drive to work alone
