Answer:
47.54% probability that more than 20 households but fewer than 35 households receive a retirement income
Step-by-step explanation:
We use the binomial aproxiation to the normal 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:
. So
![\mu = E(X) = np = 120*0.173 = 20.76](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%20120%2A0.173%20%3D%2020.76)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.173*0.827} = 4.14](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%7B120%2A0.173%2A0.827%7D%20%3D%204.14)
In a random sample of 120 households, what is the probability that more than 20 households but fewer than 35 households receive a retirement income?
We are working with discrete values, so this is the pvalue of Z when X = 35-1 = 34 subtracted by the pvalue of Z when X = 20 + 1 = 21.
X = 34
![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{34 - 20.76}{4.14}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B34%20-%2020.76%7D%7B4.14%7D)
![Z = 3.2](https://tex.z-dn.net/?f=Z%20%3D%203.2)
has a pvalue of 0.9993
X = 21
![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{21 - 20.76}{4.14}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B21%20-%2020.76%7D%7B4.14%7D)
![Z = 0.06](https://tex.z-dn.net/?f=Z%20%3D%200.06)
has a pvalue of 0.5239
0.9993 - 0.5239 = 0.4754
47.54% probability that more than 20 households but fewer than 35 households receive a retirement income