Answer:
38.76% probability that between 13 and 17 of these young adults lived with their parents
Step-by-step explanation:
I am going to use the normal approxiation 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:
![E(X) = np](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
,
.
In this problem, we have that:
![p = 0.142, n = 125](https://tex.z-dn.net/?f=p%20%3D%200.142%2C%20n%20%3D%20125)
So
![\mu = E(X) = np = 125*0.142 = 17.75](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%20125%2A0.142%20%3D%2017.75)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{125*0.142*0.858} = 3.9025](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B125%2A0.142%2A0.858%7D%20%3D%203.9025)
What is the probability that between 13 and 17 of these young adults lived with their parents?
Using continuity correction, this is
, which is the pvalue of Z when X = 17.5 subtracted by the pvalue of Z when X = 12.5. So
X = 17.5
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{17.5 - 17.75}{3.9025}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B17.5%20-%2017.75%7D%7B3.9025%7D)
![Z = -0.06](https://tex.z-dn.net/?f=Z%20%3D%20-0.06)
has a pvalue of 0.4761
X = 12.5
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{12.5 - 17.75}{3.9025}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B12.5%20-%2017.75%7D%7B3.9025%7D)
![Z = -1.35](https://tex.z-dn.net/?f=Z%20%3D%20-1.35)
has a pvalue of 0.0885
0.4761 - 0.0885 = 0.3876
38.76% probability that between 13 and 17 of these young adults lived with their parents