Question

the average weekly food expenditure of a group of families has normal distribution with mean 1800 and standard deviation 300 what is probability that out of 5 families belonging to this group at least one family has weekly food expenditure

Answer 1

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 $0.5^{5}$.

Therefore the probability of atleast one (i.e.

one or more) family with expenditure greater

than mean is 1 - $0.5^{5}$= 0.96875

Answer 2

Answer:

0.96875

Step-by-step explanation:

Since the given distribution is normal,

The probability of expenditure = Less than the Mean = 0.5

Thus, Probability of all 5 families is less than the mean = 0.5⁵

Thus, Probability of at least one family has weekly food expenditure = 1 - 0.5⁵ = 1 - 0.03125 = 0.96875

Hence, the required probability = 0.96875