Respuesta :
Answer:
There is an 1.39% probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone.
Step-by-step explanation:
For each American, there are only two possible outcomes. Either they indicate that the average person is not very considerate of others when talking on a cellphone, of they indicate that the average person is considerate. This means that we use the binomial probability distribution to solve this problem.
This distribution 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), \sigma = \sqrt{V(X)}[/tex]
In this problem, we have that:
There are 100 americans, so [tex]n = 100[/tex]
51% of Americans say the average person is not very considerate of others when talking on a cellphone. This means that [tex]p = 0.51[/tex].
Find the approximate probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone.
This is 1 subtracted by the pvalue of Z when [tex]X = 62[/tex].
We have that
[tex]\mu = E(X) = np = 100*0.51 = 51[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.51*0.49} = 5[/tex]
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{62 - 51}{5}[/tex]
[tex]Z = 2.2[/tex]
[tex]Z = 2.2[/tex] has a pvalue of 0.9861.
This means that there is a 1-0.9861 = 0.0139 = 1.39% probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone.
