Question

Suppose that a real estate negotiator has ten potential house buyers. The negotiator believes that for each potential house buyer, the probability of making a sale is 0.50.

(a) Find the probability that the negotiator makes less than two sales.

(b) Find the probability that the negotiator makes between eight and ten sales (inclusive).

(c) Find the mean and standard deviation of sales that the negotiator will be able to make.

(d) The negotiator can use a normal approximation for this binomial distribution. Explain why this is possible.

(e) Using normal approximation, calculate the probabilities in (a) and (b). [12]

Answer 1

Solution

(a):

$p=0.5,q=0.5,n=10$

$X\sim Bin(n,p)$

$P(X<2)=P(X=0)+P(X=1) \\=^{10}C_0(0.5)^{0}(0.5)^{10}+^{10}C_1(0.5)^1(0.5)^9 \\=(0.5)^{10}+10(0.5)^{10}=0.010742$

(b):

$P(8\le X\le 10)=P(X =8)+P(X =9)+P(X=10) \\=^{10}C_8(0.5)^{8}(0.5)^{2}+^{10}C_9(0.5)^9(0.5)^1+^{10}C_{10}(0.5)^{10}(0.5)^{0}$

$=45(0.5)^{10}+10(0.5)^{10}+(0.5)^{10} \\=0.0546875$

(c):

Mean$=np=10(0.5)=5$

Standard deviation$=\sqrt{npq}=\sqrt{10(0.5)(0.5)}=1.581138$

(d):

A normal approximation can be used for this binomial distribution as here $p=1/2$

(e):

Now, $X\sim N(\mu,\sigma)$

Here, $\mu=np=5, \sigma=\sqrt{npq}=1.581138$

So, $P(X<2)=P(z<\dfrac{2-5}{1.581138})$

$=P(z<-1.90)=0.02872$

Also, $P(8\le X\le 10)=P(\dfrac{8-5}{1.581138}\le z \dfrac{10-5}{1.581138} \le10)$

$=P(1.90\le z \le 3.16) \\=P(z\le3.16)-P(Z\le 1.90) \\=0.99921-0.97128 \\=0.02793$