The average age of residents in a large residential retirement community is 69 years with standard deviation 5.8 years The probability that the average age, xbar of the 100 residents selected is less than 68.5 years is:

The probability that the average  of the 100 residents selected is less than 68.5 years is

P( x⁻< 68.5) = 0.1949

Step-by-step explanation:

Step(i):-

Given mean of the Population = 69

Given standard deviation of the Population = 5.8 years

Let x⁻ be the average random variable in normal distribution

Given     x⁻  = 68.5

By using

              Z   = -0.862

Step(ii):-

The probability that the average  of the 100 residents selected is less than 68.5 years is

P( x⁻< 68.5) = P( z <-0.862)

                   =  1- P( Z >- 0.862)

                   =   1 - ( 0.5 + A(-0.862)

                  =    0.5 - A( 0.862)              (  A( -Z) = A(Z))

                  =   0.5 - 0.3051

                 =  0.1949

Conclusion:-

The percentage of the probability that the average  of the 100 residents selected is less than 68.5 years  = 19 %

three-fourths of an hour

36÷48=0.75

0.75=