Suppose that two teams play a series of games that ends when one of them has won games. Also suppose that each game played is won by team with probability p, independent of all other games. Find the expected number of games that are played when

(a) E(X) = -2p² + 2p + 2; d²/dp² E(X) at p = 1/2 is less than 0

(b) 6p⁴ - 12p³ + 3p² + 3p + 3; d²/dp² E(X) at p = 1/2 is less than 0

Step-by-step explanation:

(a) when i = 2, the expected number of played games will be:

E(X) = 2[p² + (1-p)²] + 3[2p² (1-p) + 2p(1-p)²] = 2[p²+1-2p+p²] + 3[2p²-2p³+2p(1-2p+p²)] = 2[2p²-2p+1] + 3[2p² - 2p³+2p-4p²+2p³] =  4p²-4p+2-6p²+6p = -2p²+2p+2.

If p = 1/2, then:

d²/dp² E(X) = d/dp (-4p + 2) = -4 which is less than 0. Therefore, the E(X) is maximized.

(b) when i = 3;

E(X) = 3[p³ + (1-p)³] + 4[3p³(1-p) + 3p(1-p)³] + 5[6p³(1-p)² + 6p²(1-p)³]

Simplification and rearrangement lead to:

E(X) = 6p⁴-12p³+3p²+3p+3

if p = 1/2, then:

d²/dp² E(X) at p = 1/2 = d/dp (24p³-36p²+6p+3) = 72p²-72p+6 = 72(1/2)² - 72(1/2) +6 = 18 - 36 +8 = -10

Therefore, E(X) is maximized.

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;

2

step-by-step explanation: