Question

Assume that 20.5% of all cars in South Africa are silver in colour. Let bp be the sample proportion

of silver cars in a random sample of 1000 cars.

(a) Find the mean and standard deviation of bp: (6)

(b) What is the probability that the sample proportion bp is more than 25%? (6)

(c) What is the probability that the sample proportion bp is within 1% of the true population pro-

portion? (8)

Answer 1

Solution:

n=1000, $p=20.45\% =0.2045$

$X\sim Bin(n,p)$

(a) Mean$=p=0.2045$ , Standard deviation $=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.2045(1-0.2045)}{1000}}=0.01275$

(b) $P(X>0.25)=1-P(X\le0.25)$

$=1-P(z<\dfrac{0.25-0.2045}{0.01275})$

$=1-P(z<3.56) \\=1-0.99981 \\=0.00019$

(c) We know that population proportion is 0.205(rounding off 0.2045) and hence 1% limit is (0.20295,0.20705). Taking SD = 0.0128 (rounding off 0.01275).

$P(0.20295<bp<0.20705)=P(\dfrac{0.20295-0.205}{0.0128}<z< \dfrac{0.20705-0.205}{0.0128}) \\=P(-0.16<z< 0.16) \\=P(z<0.16)-P(z<-0.16) \\=P(z<0.16)-1+P(z<0.16) \\=2\times 0.56356-1 \\=0.12712$