Answer to Question #88422 in Statistics and Probability for Penelope Mahlangu

Question

It is estimated that 30% of all drivers have some kind of medical aid in South Africa. What is the
probability that in a sample of 10 drivers:
3.1.1 Exactly 4 will have a medical aid. (8)
3.1.2 At least 2 will have a medical aid. (8)
3.1.3 More than 9 will have a medical aid. (9)
3.2 Sun Couriers, a parcel delivery company, has found that the delivery time of parcels to clients in
the Durban metropolitan area after airport collection is normally distributed with a mean
delivery time equal 45minutes (µ = 45) and a standard deviation of 8 minutes (α=8).
For a newly arrived consignment at Durban airport, what is the probability that a randomly selected
parcel will take: Between 45 and 51 minutes to deliver to the client (10)

Answer 1

The binomial distribution is used when there are exactly two mutually exclusive outcomes of a trial.The binomial distribution is used to obtain the probability of observing x successes in n trials, with the probability of success on a single trial denoted by p. The binomial distribution assumes that p is fixed for all trials. The formula for the binomial probability mass function is

$P(X=x)=\begin{pmatrix} n \\ x \end{pmatrix}p^x(1-p)^{n-x}$

3.1.1 Exactly 4 will have a medical aid.

$P(X=4)=\begin{pmatrix} 10 \\ 4 \end{pmatrix}0.3^4(1-0.3)^{10-3}=0.200120949$

3.1.2 At least 2 will have a medical aid. 

$P(X\geq2)=1-P(X<2)=1-P(X=0)-P(X=1)=$

$=1-\begin{pmatrix} 10 \\ 0 \end{pmatrix}0.3^0(1-0.3)^{10-0}-\begin{pmatrix} 10 \\ 1 \end{pmatrix}0.3^{10}(1-0.3)^{10-10}=0.850692$

3.1.3 More than 9 will have a medical aid. 

$P(X>9)=P(X=10)=\begin{pmatrix} 10 \\ 10 \end{pmatrix}0.3^{10}(1-0.3)^{10-10}=0.000006$

3.2 

$\mu=45, \sigma=8$

$z={x-\mu \over \sigma }$

$z_1={45-45 \over 8 }=0, z_2={51-45 \over 8 }=0.75$

$P(45<X<51)=P(0<Z<0.75)=0.273373$