Respuesta :
Answer:
68.29% probability that between 20 and 30 of these shoppers only review one page when searching online
Step-by-step explanation:
I am going to use the normal approximation to the binomial 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 = 100, p = 0.28[/tex]
So
[tex]\mu = E(X) = np = 100*0.28 = 28[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.28*0.72} = 4.49[/tex]
What is the probability that between 20 and 30 of these shoppers only review one page when searching online?
Using continuity correction, this is [tex]P(20 - 0.5 \leq X \leq 30 + 0.5) = P(19.5 \leq X \leq 30.5)[/tex], which is the pvalue of Z when X = 30.5 subtracted by the pvalue of Z when X = 19.5.
X = 30.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{30.5 - 28}{4.49}[/tex]
[tex]Z = 0.56[/tex]
[tex]Z = 0.56[/tex] has a pvalue of 0.7123.
X = 19.5
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{19.5 - 28}{4.49}[/tex]
[tex]Z = -1.89[/tex]
[tex]Z = -1.89[/tex] has a pvalue of 0.0294
0.7123 - 0.0294 = 0.6829
68.29% probability that between 20 and 30 of these shoppers only review one page when searching online
