Question

A carnival has a ring-toss game where players try to toss rings around a stick. A player gets two attempts in a game. Let X represent the number of rings a randomly chosen customer successfully tosses around the stick in a game. Based on previous data, here's the probability distribution of X along with summary statistics:

X=\text{\# of successes}X=# of successesX, equals, start text, \#, space, o, f, space, s, u, c, c, e, s, s, e, s, end text 000 111 222 P(X)P(X)P, left parenthesis, X, right parenthesis 0.900.900, point, 90 0.080.080, point, 08 0.020.020, point, 02 Mean: \mu_X=0.12μ X ​ =0.12mu, start subscript, X, end subscript, equals, 0, point, 12 Standard deviation: \sigma_X\approx0.38σ X ​ ≈0.38sigma, start subscript, X, end subscript, approximately equals, 0, point, 38
A player wins 10 tickets for each successful toss. Let T represent the number of tickets a randomly chosen players wins in a game. What are the mean and standard deviation of T?

Answer 1

To find the mean and standard deviation of T, multiply the mean and variance of X by 10. The mean of T is 1.2 and the standard deviation is approximately 3.8.

Step-by-step explanation:

To find the mean and standard deviation of T, we need to calculate the expected value and variance of T. Since each successful toss earns 10 tickets, the random variable T is equal to 10 times the number of successful tosses X. Thus, T = 10X.

The mean of T, denoted as μT, is given by μT = 10μX. Since the mean of X is 0.12, the mean of T is 10(0.12) = 1.2.

The variance of T, denoted as σT2, is given by σT2 = (102)σX2. Since the standard deviation of X is 0.38, the variance of T is (102)(0.382) = 14.44. Taking the square root of the variance gives the standard deviation of T, which is approximately 3.8.

Answer 2

1.2

3.8

Step-by-step explanation:

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