Question

) 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.

Calculate the probability that

ii. FOUR faulty reports will be made on Monday

iii. Less than 3 faulty reports in a 5-day work week

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.

(ii) Given that the netballer made a total of 120 attempts at shooting goals in a tournament, calculate the expected number of successful shoots. 

Answer 1

a.

$i)$

The random variable in this question is the number of faulty reports made to the company's switchboard.

Let $X$ be a random variable representing the number of faulty reports made to the company's switchboard then, $X\sim Poisson(\lambda)$ given as,

$p(X=x)=e^{-\lambda}\lambda^x/x!\space x=0,1,2,3,..$

For this case, $\lambda=2.5$ per day

$ii)$

The probability that four faulty reports will be made on Monday is given as,

$p(X=4)=e^{-2.5}2.5^4/4!=0.1336019$

Therefore, the probability that there are four faulty reports on Monday is 0.1336.

$iii)$

The rate $\lambda=2.5$ per day, therefore in a 5-day work week, $\lambda=2.5*5=12.5$

Therefore the probability that the number of faulty reports is less than 3 is,

$p(X\lt3)=p(X=0)+p(X=1)+p(X=2)=(e^{-12.5}12.5^0/0!)+(e^{-12.5}12.5^1/1!)+(e^{-12.5}12.5^2/2!)= 3.726653e-06+ 4.658316e-05+ 0.0002911448= 0.0003414546$

The probability that there are less than 3 faulty reports in a 5-day work week is 0.0003414546

b.

$i)$

Here, the probability of success is $p=0.35$ and $n=12$. The number of success $x=4$. To find the probability that there are exactly 4 successes, we apply the binomial distribution as follows.

$p(X=4)=\binom{12}{4}0.35^40.65^8= 0.2366924$

$ii)$

Given $n=120$ attempts and $p=0.35$, the expected number of successful shoots is given by the mean of the binomial distribution given as $np.$

Therefore, the expected number of successful shoots is $E(X)=np=120*0.35=42$.