Question

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 ) to are to count the number of times the topside showed a 6. this count is represented by the random variable ( x ).
) the distribution of ( x ) with a mean and a standard deviation enter your answe
ing all the decimals you
) now think about thesses that show a six. what can you say about the distribution of this proportion? complete th
tence. enter your ans
the distribution of ( hat{p} ) with a mean ( mu_{hat{p}}= ) and a standard deviation ( sigma_{hat{p}}= )
(c) what is the probabilityge 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: Standardize the values

For \(p_1 = 0.14\) and \(p_2=0.22\), use the formula \(z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}\).
For \(p_1 = 0.14\):
\(z_1=\frac{0.14 - 0.1667}{0.0350}=\frac{- 0.0267}{0.0350}\approx - 0.763\)
For \(p_2 = 0.22\):
\(z_2=\frac{0.22 - 0.1667}{0.0350}=\frac{0.0533}{0.0350}\approx1.523\)

Step2: Find the probability

We want \(P(0.14<\hat{p}<0.22)\), which is equivalent to \(P(-0.763<Z<1.523)\)
Using the standard normal table:
\(P(Z < 1.523)\approx0.9364\)
\(P(Z<-0.763)\approx0.2236\)
\(P(-0.763 < Z < 1.523)=P(Z < 1.523)-P(Z<-0.763)\)
\(=0.9364 - 0.2236=0.7128\)

Answer:

\(0.7128\)