Answer:
a) Number of people expected to have consumed alcoholic beverages in the sample = 35.
Standard deviation = 3.24
b) 45 or more people consuming alcoholic beverages in the sample is about 3 standard deviations from the mean (35), hence, it would be surprising if this happened since most of the possible values in a distribution stay well within a few standard deviations of the mean.
c) The probability that 45 or more people in this sample have consumed alcoholic beverages = 0.0017
The mathematical and statistical chances of 45 or more people who have consumed alcoholic beverages in the sample is very low (0.0017), hence, it will be very surprising if this happened.
Step-by-step explanation:
Concluding part of the question
What is the probability that 45 or more people in this sample have consumed alcoholic beverages? How does this probability relate to your answer to part (b)?
a) The expected number of people that take alcoholic beverages in a sample of people is the mean of a binomial distribution and it is given as
Mean = np
where n = sample size = 50
p = population proportion of youths that take alcoholic beverages = 70% = 0.70
Number of people expected to have consumed alcoholic beverages in the sample = 0.70 × 50 = 35
The standard deviation is given as
σ = √[np(1-p)] = √(50×0.70×0.30) = 3.24
b and c) To investigate whether one should be surprised if 45 or more people out of the sample take alcoholic beverages, we first find the probability of this happening.
If the probability is very small or minute.
Using the normal approximation to the binomial distribution,
μ = 35
σ = 3.24
If X denotes the number of people that will take alcoholic beverages in the sample.
And we need P(X ≥ 45)
we first introduce the continuity correction factor.
P(X ≥ 45) = P(x ≥ 44.5)
We normalize or standardize the number 44.5
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (44.5 - 35)/3.24 = 2.93
To determine the required probability
We'll use data from the normal distribution table for these probabilities
P(X ≥ 45) = P(x ≥ 44.5) = P(z ≥ 2.93)
= 1 - P(z < 2.93)
= 1 - 0.99831
= 0.00169 = 0.0017 to 4 d.p.
Hope this Helps!!!
