Respuesta :
Answer:
a) 0.0548 = 5.48% probability that at least 20 % of those in the sample will be retired.
b) 0.8388 = 83.88% probability that between 15 % and 21 % of those in the sample will be retired.
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
17.0 % of mutual fund shareholders are retired persons.
This means that [tex]p = 0.17[/tex]
Assuming a simple random sample of 400 shareholders has been selected
This means that [tex]n = 400[/tex]
Mean and standard deviation:
[tex]\mu = p = 0.17[/tex]
[tex]s = \sqrt{\frac{0.17*0.83}{400}} = 0.0188[/tex]
a. What is the probability that at least 20 % of those in the sample will be retired?
This is the 1 p-value of Z when X = 0.2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.2 - 0.17}{0.0188}[/tex]
[tex]Z = 1.6[/tex]
[tex]Z = 1.6[/tex] has a p-value of 0.9452
1 - 0.9452 = 0.0548
0.0548 = 5.48% probability that at least 20 % of those in the sample will be retired.
b. What is the probability that between 15 % and 21 % of those in the sample will be retired?
This is the p-value of Z when X = 0.21 subtracted by the p-value of Z when X = 0.15.
X = 0.21
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.21 - 0.17}{0.0188}[/tex]
[tex]Z = 2.13[/tex]
[tex]Z = 2.13[/tex] has a p-value of 0.9834
X = 0.15
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.15 - 0.17}{0.0188}[/tex]
[tex]Z = -1.06[/tex]
[tex]Z = -1.06[/tex] has a p-value of 0.1446
0.9834 - 0.1446 = 0.8388
0.8388 = 83.88% probability that between 15 % and 21 % of those in the sample will be retired.
