Answer to Question #305448 in Statistics and Probability for Alak

Question

Monthly food expenditures for families in Rodriguez Rizal averaged Php6,500 w

a standard deviation of Php750. Assuming that the monthly food expenditures are

normally distributed. What percentages of these expenditures are between Pho5.900

and Php7,8502

Answer 1

In this case, we need to find the area between X = 5,900 and X =7,850

By using the Z score formula given by:

Z = $\frac{X -\mu}{\sigma}$ 

Given $\mu$ = 6,500 and $\sigma$ = 750

When X = 5,900

Z = $\frac{5,900 -6,500}{750}$  = -0.8

When X = 7,850

Z = $\frac{7,850 -6,500}{750}$  = 1.8

Therefore, we need to find the area between Z = -0.8 and Z = 1.8

That is;

P(-0.8<Z<1.8) = P(Z<1.8) - P(Z<-0.8)

From the standard normal table,

P(Z<1.8) = 0.9641

Also,

P(Z<-0.8) = 0.2119

Therefore,

P(-0.8<Z<1.8) = P(Z<1.8) - P(Z<-0.8) = (0.9641 - 0.2119) = 0.7522

Answer: 0.7522