Question

(1 point)
you are to roll a fair die ( n = 113 ) times, each time observing if the topside of the die shows a 6 (success) or not (failure). after observing the ( n = 113 ) tosses, you are to count the number of times the topside showed a 6. this count is represented by the random variable ( x )
(a) the distribution of ( x ) is approximately normal with a mean and a standard deviation enter your answer using all the decimals you can
(b) now think about the proportion of your ( n = 113 ) tosses that show a six. what can you say about the distribution of this proportion? complete the sentence. enter your answer using all the decimals you can.
the distribution of ( hat{p} ) is approximately normal with a mean ( mu_{hat{p}}=) and a standard deviation ( sigma_{hat{p}}=)
(c) what is the probability that the proportion/percentage of your ( n = 113 ) tosses that show a six will be somewhere between ( 14 % ) and ( 22 % )? enter your answer using all the decimals you can.
(d) after the ( n = 113 ) tosses of the die, you observe ( x = 25 ), the value of the sample proportion is then ( hat{p}=\frac{25}{113}=0.2212 ). what is the probability of observing a sample proportion that is at least this much should you decide to roll this die again 113 times? enter your answer using all the decimals you can.

Explanation:

Step1: Calculate the mean and standard deviation for \(X\)

For a binomial distribution \(X\sim B(n,p)\), the mean \(\mu = np\) and the standard deviation \(\sigma=\sqrt{np(1 - p)}\). Here, \(n = 113\) and \(p=\frac{1}{6}\) (probability of getting a 6 on a fair die).

• Mean of \(X\): \(\mu=np=113\times\frac{1}{6}\approx18.8333\)
• Standard deviation of \(X\): \(\sigma=\sqrt{np(1 - p)}=\sqrt{113\times\frac{1}{6}\times(1-\frac{1}{6})}=\sqrt{\frac{113}{6}\times\frac{5}{6}}=\sqrt{\frac{565}{36}}\approx3.9613\)

Step2: Calculate the mean and standard deviation for \(\hat{p}\)

For the sample proportion \(\hat{p}\), the mean \(\mu_{\hat{p}}=p\) and the standard deviation \(\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}\).

• Mean of \(\hat{p}\): \(\mu_{\hat{p}}=\frac{1}{6}\approx0.1667\)
• Standard deviation of \(\hat{p}\): \(\sigma_{\hat{p}}=\sqrt{\frac{\frac{1}{6}(1-\frac{1}{6})}{113}}=\sqrt{\frac{\frac{5}{36}}{113}}=\sqrt{\frac{5}{4068}}\approx0.0350\)

Step3: Calculate the probability for part (c)

First, convert the proportions to \(z\)-scores. For \(p_1 = 0.14\) and \(p_2=0.22\)

• \(z_1=\frac{0.14 - 0.1667}{0.0350}=\frac{- 0.0267}{0.0350}\approx - 0.7629\)
• \(z_2=\frac{0.22 - 0.1667}{0.0350}=\frac{0.0533}{0.0350}\approx1.5229\)

Using the standard normal distribution \(P(-0.7629<Z<1.5229)=\Phi(1.5229)-\Phi(-0.7629)\)
\(\Phi(1.5229)\approx0.9364\), \(\Phi(-0.7629)\approx0.2224\)
\(P(-0.7629<Z<1.5229)=0.9364 - 0.2224 = 0.7140\)

Step4: Calculate the probability for part (d)

For \(\hat{p}=0.2212\), \(z=\frac{0.2212 - 0.1667}{0.0350}=\frac{0.0545}{0.0350}\approx1.5571\)
\(P(Z\geq1.5571)=1-\Phi(1.5571)\)
\(\Phi(1.5571)\approx0.9409\)
\(P(Z\geq1.5571)=1 - 0.9409=0.0591\)

Answer:

• (a) Mean: \(18.8333\), Standard deviation: \(3.9613\)
• (b) Mean \(\mu_{\hat{p}}:0.1667\), Standard deviation \(\sigma_{\hat{p}}:0.0350\)
• (c) \(0.7140\)
• (d) \(0.0591\)