Answer to Question #273332 in Statistics and Probability for Light

Question

professional basketball player makes 85% of the free throws he tries. Assuming this percentage will hold true for all future attempts, find the probability that in the next eight attempts, the number of free throws he will make is

i. exactly 8 [4 marks]

ii. less than 3 [4 marks]

b) A survey indicates that people use their cellular phones an average of 1.5 years before buying a new one.The standard deviation is 0.25 year. A cellular phone user is selected at random. Find the probability that the user will use their current phone for less than 1 year before buying a new one.[ Assume that the times ( X years) are normally distrbuted]. [4 marks]

Answer 1

a) Let X be the number of free throws he will make in the next 8 tries.

He makes 85% of the free throws he tries.

X~Bin(8,0.85)

$P(X) = C^{8}_x(0.85)^x(1-0.85)^{8-x}$

(i) exactly 8

$P(X=8) = C^8_8 (0.85)^8(1-0.85)^{8-8} = 0.85^8 = 0.27249$

(ii) less than 3

$P(X<3) = P(X=0) + P(X=1) + P(X=2) \\ P(X=0) = C^{8}_0(0.85)^0(1-0.85)^{8-0} = 0.15^8 = 2.56 \times 10^{-7} \\ P(X=1) = C^{8}_1(0.85)^1(1-0.85)^{8-1} = 1.16 \times 10^{-5} \\ P(X=2) = C^{8}_2(0.85)^2(1-0.85)^{8-2} = 0.0002304 \\ P(X<3) = 2.56 \times 10^{-7} + 1.16 \times 10^{-5} + 0.0002304 = 0.000242289$

b)

$\mu = 1.5 \\ \sigma = 0.25 \\ P(X<1) = P(Z< \frac{1-1.5}{0.25}) \\ = P(Z< -2) \\ = 0.0228$