Question

The average weekly expenditure on a particular commodity incurred by families in a certain locality is normally distributed with a mean of Rs.125 and a standard deviation equal to Rs.25. What is the probability that a family selected at random from the locality, will have an average weekly expenditure on the given commodity in excess of Rs.175 ? What is the probability that out of eight families selected from the locality, at least one family will incur an average weekly expenditure on the given commodity in excess of Rs.175 ?

Answer 1

Answer:

Answer:

Step-by-step explanation:

Given that the distribution is normal

(actually it doesn't matter what thet

distribution is), the probability that the

expenditure of a family is less than the mean,

by definition, is 0.5. Since the expenditure of

a family is independent of the expenditures

of other families, the probability that the

expenditure of all the 5 families is less than

the mean is 0.5^{5}0.5

5

.

Therefore the probability of atleast one (i.e.

one or more) family with expenditure greater

than mean is 1 - 0.5^{5}0.5

5

= 0.96875

Answer 2

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