Respuesta :
Answer:
a) 35.94% probability that 160 or fewer find their jobs stressful.
b) 98.46% probability that more than 150 find their jobs stressful.
c) 39.62% probability that the number who find their jobs stressful is between 155 and 162, inclusive.
Step-by-step explanation:
We 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]n = 200, p = 0.81[/tex]. So
[tex]\mu = E(X) = np = 200*0.81 = 162[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.81*0.19} = 5.55[/tex]
a. Approximate the probability that 160 or fewer find their jobs stressful.
This is the pvalue of Z when X = 160. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{160 - 162}{5.55}[/tex]
[tex]Z = -0.36[/tex]
[tex]Z = -0.36[/tex] has a pvalue of 0.3594.
35.94% probability that 160 or fewer find their jobs stressful.
b. Approximate the probability that more than 150 find their jobs stressful.
This is 1 subtracted by the pvalue of Z when X = 150. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{150 - 162}{5.55}[/tex]
[tex]Z = -2.16[/tex]
[tex]Z = -2.16[/tex] has a pvalue of 0.0154.
1 - 0.0154 = 0.9846
98.46% probability that more than 150 find their jobs stressful.
c. Approximate the probability that the number who find their jobs stressful is between 155 and 162, inclusive.
This is the pvalue of Z when X = 162 subtracted by the pvalue of Z when X = 155. So
X = 162
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{162 - 162}{5.55}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a pvalue of 0.5
X = 155
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{155 - 162}{5.55}[/tex]
[tex]Z = -1.26[/tex]
[tex]Z = -1.26[/tex] has a pvalue of 0.1038
0.5 - 0.1038 = 0.3962
39.62% probability that the number who find their jobs stressful is between 155 and 162, inclusive.
