Answer:
(a) <em> </em><em>n</em> : 20 50 100 500
P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886 0.4444 0.5954 0.9376
(b) The correct option is (b).
Step-by-step explanation:
Let the random variable <em>X</em> represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.
The mean amount of deductions is, <em>μ</em> = $16,642 and standard deviation is, <em>σ</em> = $2,400.
Assuming that the random variable <em>X </em>follows a normal distribution.
(a)
Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:
- For a sample size of <em>n</em> = 20
![P(\mu-200](https://tex.z-dn.net/?f=P%28%5Cmu-200%3CX%3C%5Cmu%2B200%29%3DP%28%5Cfrac%7B%2816642-200%29-16642%7D%7B2400%2F%5Csqrt%7B20%7D%7D%3C%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3C%5Cfrac%7B%2816642%2B200%29-16642%7D%7B2400%2F%5Csqrt%7B20%7D%7D%29)
![=P(-0.37](https://tex.z-dn.net/?f=%3DP%28-0.37%3CZ%3C0.37%29%5C%5C%5C%5C%3DP%28Z%3C0.37%29-P%28Z%3C-0.37%29%5C%5C%5C%5C%3D0.6443-0.3557%5C%5C%5C%5C%3D0.2886)
- For a sample size of <em>n</em> = 50
![P(\mu-200](https://tex.z-dn.net/?f=P%28%5Cmu-200%3CX%3C%5Cmu%2B200%29%3DP%28%5Cfrac%7B%2816642-200%29-16642%7D%7B2400%2F%5Csqrt%7B50%7D%7D%3C%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3C%5Cfrac%7B%2816642%2B200%29-16642%7D%7B2400%2F%5Csqrt%7B50%7D%7D%29)
![=P(-0.59](https://tex.z-dn.net/?f=%3DP%28-0.59%3CZ%3C0.59%29%5C%5C%5C%5C%3DP%28Z%3C0.59%29-P%28Z%3C-0.59%29%5C%5C%5C%5C%3D0.7222-0.2778%5C%5C%5C%5C%3D0.4444)
- For a sample size of <em>n</em> = 100
![P(\mu-200](https://tex.z-dn.net/?f=P%28%5Cmu-200%3CX%3C%5Cmu%2B200%29%3DP%28%5Cfrac%7B%2816642-200%29-16642%7D%7B2400%2F%5Csqrt%7B100%7D%7D%3C%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3C%5Cfrac%7B%2816642%2B200%29-16642%7D%7B2400%2F%5Csqrt%7B100%7D%7D%29)
![=P(-0.83](https://tex.z-dn.net/?f=%3DP%28-0.83%3CZ%3C0.83%29%5C%5C%5C%5C%3DP%28Z%3C0.83%29-P%28Z%3C-0.83%29%5C%5C%5C%5C%3D0.7977-0.2023%5C%5C%5C%5C%3D0.5954)
- For a sample size of <em>n</em> = 500
![P(\mu-200](https://tex.z-dn.net/?f=P%28%5Cmu-200%3CX%3C%5Cmu%2B200%29%3DP%28%5Cfrac%7B%2816642-200%29-16642%7D%7B2400%2F%5Csqrt%7B500%7D%7D%3C%5Cfrac%7BX-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3C%5Cfrac%7B%2816642%2B200%29-16642%7D%7B2400%2F%5Csqrt%7B500%7D%7D%29)
![=P(-1.86](https://tex.z-dn.net/?f=%3DP%28-1.86%3CZ%3C1.86%29%5C%5C%5C%5C%3DP%28Z%3C1.86%29-P%28Z%3C-1.86%29%5C%5C%5C%5C%3D0.9688-0.0312%5C%5C%5C%5C%3D0.9376)
<em> n</em> : 20 50 100 500
P (-200 < <em>X</em> - <em>μ </em>< 200) : 0.2886 0.4444 0.5954 0.9376
(b)
The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample (
) approaches the whole population mean (
).
Consider the probabilities computed in part (a).
As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.
So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.
Thus, the correct option is (b).