Answer to Question #334000 in Statistics and Probability for Hansi

Question

The average amount parents spent per child on private tuition classes is Rs 6000.

Assume that the standard deviation is Rs 1000 and that the amount spent is normally

distributed.

(i) Find the probability that the amount spent on a randomly selected child is:

a. Less than Rs 5000

b. Between Rs 5000 and Rs 6500

c. What is the probability that the amount spent on a randomly selected

child is between Rs 6500 and RS 8000. (06 marks)

(ii) Assume that 5% of parents spent over Rs 8000. If we selected randomly 1000

parents who sent their children to tuition, how many parents would spend more

than Rs 8000?

Answer 1

Let $X=$ the amount spent: $X\sim N(\mu, \sigma^2).$

(i) Given $\mu=6000\ Rs, \sigma=1000\ Rs.$

a.

$$P(X<5000)=P(Z<\dfrac{5000-6000}{1000})$$

$$=P(Z<-1)\approx0.158655$$

b.

$$P(5000<X<6500)=P(Z<\dfrac{6500-6000}{1000})$$

$$-P(Z\le\dfrac{5000-6000}{1000})$$

$$=P(Z<0.5)-P(Z\le-1)$$

$$\approx0.691462-0.158655\approx0.5328$$

c.

$$P(6500<X<8000)=P(Z<\dfrac{8000-6000}{1000})$$

$$-P(Z\le\dfrac{6500-6000}{1000})$$

$$=P(Z<2)-P(Z\le0.5)$$

$$\approx0.977250-0.691462\approx0.2858$$

(ii)

$$1000(0.05)=50$$

50 parents