In: Statistics and Probability
Given that a recent report in Pasadena Times indicated a typical family of four spends $490 per month on food.
Here we assume that the distribution of food expenditures for a family follows the normal distribution, with a standard deviation of $90 per month.
Let,
X=Expenditure of a typical family.
Before we go on to solve the problem let us know a bit about Normal Distribution.
Normal Distribution
A continuous random variable X is said to have a normal distribution if its PDF(Probability Density Function) is given by
its CDF(Cumulative Distribution Function) is given by,
Notation:
Standard Normal Distribution
A continuous random variable X is said to have a standard normal distribution if its PDF(Probability Density Function) is given by
its CDF(Cumulative Distribution Function) is given by,
Exact evaluation of ?(x) is not possible but numerical method can be applied. The values of ?(x) has been tabulated extensively in Biometrika Volume I.
Notation:
Note:
Coming back to our problem
Let,
X=Expenditure of a typical family.
(a) Here we need to find the percent of the families who spend between $300 and $490 per month on food.
Hence (0.4826*100)%=48.26% of the families spend between $300 and $490 per month on food.
(b) Here we need to find the probability that a family selected spends less than $430 per month on food.
Hence the probability that a family selected spends less than $430 per month on food is 0.2514
(c) Here we need to find the percent of the families who spend between $430 and $600 per month on food.
Hence (0.6374*100)%=63.74% of the families spend between $430 and $600 per month on food.
(d) Here we need to find the probability of selecting a family that spends between $400 and $600 per month on food.
Hence the probability of selecting a family that spends between $400 and $600 per month on food is 0.7301