Question
A binomial event has \(n=60\) trials. The probability of success on each trial is 0.4 . Let \(X\) be the count of successes of the event during the 60 trials. What are \(\mu_{\mathrm{x}}\) and \(\sigma_{\mathrm{x}}\) ?
(a) \(24,3.79\)
(b) \(24,14.4\)
(c) \(4.90,3.79\)
(d) \(4.90,14.4\)
(e) \(2.4,3.79\)
▶️Answer/Explanation
The correct answer is (a).
$
\mu_X=(60)(0.4)=24, \sigma_X=\sqrt{60(0.4)(0.6)}=\sqrt{14.4}=3.79 \text {. }
$
Question
Consider repeated trials of a binomial random variable. Suppose the probability of the first success occurring on the second trial is 0.25 . What is the probability of success on the first trial?
(a) \(1 / 4\)
(b) 1
(c) \(1 / 2\)
(d) \(1 / 8\)
(e) \(3 / 16\)
▶️Answer/Explanation
The correct answer is (c).
If it is a binomial random variable, the probability of success, \(p\), is the same on each trial. The probability of not succeeding on the first trial and then succeeding on the second trial is \((1-p)(p)\). Thus, \((1-p) p=\) 0.25 . Solving algebraically, \(p=1 / 2\).
Question
To use a normal approximation to the binomial, which of the following does not have to be true?
(a) \(n p \geq 10, n(1-p) \geq 10\) (or: \(n p \geq 5, n(1-p) \geq 5\) ).
(b) The individual trials must be independent.
(c) The sample size in the problem must be too large to permit doing the problem on a calculator.
(d) For the binomial, the population size must be at least 10 times as large as the sample size.
(e) All of the above are true.
▶️Answer/Explanation
The answer is (b).
$
\mu_{\bar{x}}=\mu, \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} .
$
For small samples, the shape of the sampling distribution of \(\bar{x}\) will resemble the shape of the sampling distribution of the original population. The shape of the sampling distribution of \(\bar{x}\) is approximately normal for \(n\) sufficiently large.
The correct answer is (c). Although you probably wouldn’t need to use a normal approximation to the binomial for small sample sizes, there is no reason (except perhaps accuracy) that you couldn’t.
Question
You form a distribution of the means of all samples of size 9 drawn from an infinite population that is skewed to the left (like the scores on an easy Stats quiz!). The population from which the samples are drawn has a mean of 50 and a standard deviation of 12 . Which one of the following statements is true of this distribution?
(a) \(\mu_{\bar{x}}=50, \sigma_{\bar{x}}=12\), the sampling distribution is skewed somewhat to the left.
(b) \(\mu_{\bar{x}}=50, \sigma_{\bar{x}}=4\), the sampling distribution is skewed somewhat to the left.
(c) \(\mu_{\bar{x}}=50, \sigma_{\bar{x}}=12\), the sampling distribution is approximately normal.
(d) \(\mu_{\bar{x}}=50, \sigma_{\bar{x}}=4\), the sampling distribution is approximately normal.
(e) \(\mu_{\bar{x}}=50, \sigma_{\bar{x}}=4\), the sample size is too small to make any statements about the shape of the sampling distribution.
▶️Answer/Explanation
Ans:The correct answer is (d).
Because the problem stated “at least 10 ,” we must include the term where \(x=10\). If the problem had said “more than 10 ,” the correct answer would have been (b) or (c) (they are equivalent). The answer could also have been given as
Question
A 12-sided die has faces numbered from 1-12. Assuming the die is fair (that is, each face is equally likely to appear each time), which of the following would give the exact probability of getting at least 103 s out of 50 rolls?
(a) \(\left(\begin{array}{c}50 \\ 0\end{array}\right)(0.083)^0(0.917)^{50}+\left(\begin{array}{c}50 \\ 1\end{array}\right)(0.083)^1(0.917)^{49}+\ldots+\left(\begin{array}{c}50 \\ 9\end{array}\right)(0.083)^9(0.917)^{41}\).
(b) \(\left(\begin{array}{l}50 \\ 11\end{array}\right)(0.083)^{11}(0.917)^{39}+\left(\begin{array}{c}50 \\ 12\end{array}\right)(0.083)^{12}(0.917)^{38}+\ldots+\left(\begin{array}{c}50 \\ 50\end{array}\right)(0.083)^{50}(0.917)^0\).
(c) \(1-\left[\left(\begin{array}{c}50 \\ 0\end{array}\right)(0.083)^0(0.917)^{50}+\left(\begin{array}{c}50 \\ 1\end{array}\right)(0.083)^1(0.917)^{49}+\ldots+\left(\begin{array}{c}50 \\ 10\end{array}\right)(0.083)^{10}(0.917)^{40}\right]\).
(d) \(1-\left[\left(\begin{array}{c}50 \\ 0\end{array}\right)(0.083)^0(0.917)^{50}+\left(\begin{array}{c}50 \\ 1\end{array}\right)(0.083)^1(0.917)^{49}+\ldots+\left(\begin{array}{c}50 \\ 9\end{array}\right)(0.083)^9(0.917)^{41}\right]\).
(e) \(\left(\begin{array}{c}50 \\ 0\end{array}\right)(0.083)^0(0.917)^{50}+\left(\begin{array}{c}50 \\ 1\end{array}\right)(0.083)^1(0.917)^{49}+\ldots+\left(\begin{array}{c}50 \\ 10\end{array}\right)(0.083)^{10}(0.917)^{40}\).
▶️Answer/Explanation
Ans:The correct answer is(d).
$
\mu_{\hat{p}}=p=0.55, \sigma_{\hat{p}}=\sqrt{\frac{(0.55)(0.45)}{100}}=0.0497 .
$