Question

in the game of roulette, a wheel consists of 38 slots numbered 0, 00, 1, 2, ..., 36. to play the game, a metal ball is spun around the wheel and is allowed to fall into one of the numbered slots. if the number of the slot the ball falls into matches the number you selected, you win $35, otherwise you lose $1. complete parts (a) through (g) below
click here to view the standard normal distribution table (page 1).
click here to view the standard normal distribution table (page 2).
what are the mean and standard deviation of the sampling distribution of x? round your results to the nearest penny.
\\( \mu _ { \overline { x } } = - 0.05 \\)
\\( \sigma _ { \overline { x } } = 0.61 \\)
(d) what is the probability of being ahead after playing the game 90 times? that is, what is the probability that the sample mean is greater than 0 for \\( n = 90 \\)?
\\( p ( \overline { x } > 0 ) = 0.4681 \\)
(type an integer or decimal rounded to four decimal places as needed.)
(e) what is the probability of being ahead after playing the game 180 times?
\\( p ( \overline { x } > 0 ) = 0.4512 \\)
(type an integer or decimal rounded to four decimal places as needed.)
(f) what is the probability of being ahead after playing the game 900 times?
\\( p ( \overline { x } > 0 ) = \square \\)
(type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\). Here, \(\bar{x} = 0\), \(\mu_{\bar{x}}=- 0.05\), and \(n = 900\). First, we need to find \(\sigma_{\bar{x}}\) for \(n = 900\). Since \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\) (assuming the population standard deviation \(\sigma\) is known from previous parts, and here we use the property of sampling distribution of the sample mean). Given \(\sigma_{\bar{x}}\) for \(n\) (from previous, general formula for sampling distribution of the sample mean: \(\mu_{\bar{x}}=\mu\) and \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\)).

We know that \(\sigma_{\bar{x}}\) (for \(n = 900\)): \(\sigma_{\bar{x}}=\frac{0.61}{\sqrt{900}}=\frac{0.61}{30}\approx0.0203\)

The z - score is \(z=\frac{0 - (-0.05)}{0.0203}=\frac{0.05}{0.0203}\approx2.46\)

Step2: Find the probability using the standard normal distribution

We want to find \(P(\bar{X}>0)\), which is equivalent to \(P(Z > 2.46)\) (using the standard normal transformation \(Z=\frac{\bar{X}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}\)).

Since \(P(Z>z)=1 - P(Z\leq z)\), and from the standard normal table \(P(Z\leq2.46)=0.9931\)

So \(P(Z > 2.46)=1 - 0.9931=0.0069\)

Answer:

\(0.0069\)