Answer to Question #260950 in Statistics and Probability for bri

Question

a) On average 2.5 faulty reports are made to a company’s switchboard per day. i. Name the random variable present in this problem and state its distribution. [2] Calculate the probability that ii. FOUR faulty reports will be made on Monday [2] iii. Less than 3 faulty reports in a 5-day work week [4] b) The number of attempts at shooting goals made by a netballer in a tournament can be modelled by a binomial distribution with a probability of success equal to 0.35. (i) In a sample of 12 attempts at shooting goals, calculate the probability that EXACTLY 4 were successful. [4] (ii) Given that the netballer made a total of 120 attempts at shooting goals in a tournament, calculate the expected number of successful shoots. [2]

Answer 1

a)

i)

variable is the number of faulty report

Poisson distribution:

$P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!}$

mean $\lambda=2.5$ faulty reports per day

ii)

$P(x=4)=\frac{2.5^4e^{-2.5}}{4!}=0.1336$

iii)

for a 5-day work week:

$\lambda=2.5\cdot 5=12.5$

$P(x<3)=P(x=0)+P(x=1)+P(x=2)$

$P(x=0)=e^{-12.5}=3.7\cdot 10^{-6}$

$P(x=1)=12.5e^{-12.5}=4.7\cdot 10^{-5}$

$P(x=2)=\frac{12.5^2e^{-12.5}}{2}=2.9\cdot10^{-4}$

$P(x<3)=3.7\cdot 10^{-6}+4.7\cdot 10^{-5}+2.9\cdot10^{-4}=3.407\cdot10^{-4}$

b)

binomial distribution:

$P(x=k)=C^k_np^k(1-p)^{n-k}$

$p=0.35$

i)

$n=12$

$P(x=4)=C^4_{12}\cdot 0.35^4\cdot 0.65^8=0.2367$

ii)

$n=120$

the expected number of successful shoots:

$E(X)=np=120\cdot0.35=42$