Answer to Question #161699 in Statistics and Probability for Qianzi

Question

The price of all houses in Bukit Katil has a mean of RM109,000 and a standard

deviation of RM5,700. Find the probability that the mean price of a random sample of

75 selected houses would be

i. within RM2,200 of the population mean.

ii. more than the population mean by RM1,000 or more.

iii. between RM107,800 and RM110,100.

Answer 1

The price of all houses in Bukit Katil has a mean RM $109000$ and standard deviation RM $5700$ .

Let $X$ be a random variable defines mean price of $75$ selected houses

Then $\mu =109000$ and $\sigma =5700$ .

Let $Z=\frac{X-\mu}{\sigma}$ .Then $Z=\frac{X-109000}{5700}$.

(i) We have to find the probability of mean price of $75$ selected houses would be within RM $2200$ .

$\therefore X=(109000+2200)=111200$

Now $P(X<111200)=P(Z<\frac{111200-109000}{5700})$

$=P(Z<0.39)$

$=0.5+P(0<Z<0.39)$

$=0.5+0.1517$

$=0.65$ (approximately)

(ii) We have to find the probability of mean price of 75 selected houses are more than the population mean by RM $1000.$

So, $X=(109000+1000)=110000$

$\therefore P(X\geq 110000)= P(Z\geq \frac{110000-109000}{5700})$

$=P(Z\geq 0.18)$

$=0.5-P(0<Z<0.18)$

$=0.5-0.0714$

$=0.43$ (approximately)

(iii) $P(107800<X<110100)=P(\frac{107800-109000}{5700}<Z<\frac{110100-109000}{5700})$

$=P(-0.21<Z<0.19)$

$=P(0<Z<0.21)+P(0<Z<0.19)$

$=$ $0.0831+0.0753$

$=0.16$ (approximately)