Respuesta :
Answer:
0.8888 = 88.88% probability that at least 70 use their phone at least once per hour
Step-by-step explanation:
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].
51% of users of mobile phones use their phone at least once per hour.
This means that [tex]p = 0.51[/tex]
Consider a sample of 150 mobile phone users.
This means that [tex]n = 150[/tex]
So
[tex]\mu = E(X) = np = 150*0.51 = 76.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{150*0.51*0.49} = 6.12[/tex]
What is the probability that at least 70 use their phone at least once per hour?
This is 1 subtracted by the pvalue of Z when X = 70-1 = 69(at least 70 is more than 69). So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{69 - 76.5}{6.12}[/tex]
[tex]Z = -1.22[/tex]
[tex]Z = -1.22[/tex] has a pvalue of 0.1112
1 - 0.1112 = 0.8888
0.8888 = 88.88% probability that at least 70 use their phone at least once per hour
