Homework: Chapter 6 HW Score: 0 of 1 pt 25 of 30 (17 complete) 6.3.93 A variable

Question

Homework: Chapter 6 HW Score: 0 of 1 pt 25 of 30 (17 complete) 6.3.93 A variable is normally distributed with mean 15 and standard deviation 3. a. Determine the quartiles of the variable. b. Obtain and interpret the 90th percentile. . c. Find the value that 65% of all possible values of the variable exceed. d. Find the two values that divide the area under the corresponding normal curve into a middle area of 0.95 and two outside areas of 0.025. Interpret the answer Click here to view page 1 of the standard normal distribution table, a. Q1-12.99 (Round to two decimal places as needed.) 02-15 17.01 b. The 90th percentile is (Round to two decimal places as needed.)

Explanation / Answer

Solution:

Mean () =15
Standard Deviation (sd) = 3
Normal Distribution Z = X- / sd ~ N(0,1)   
a.
Q1 = P(Z < x) = 0.25
Value of z to the cumulative probability of 0.25 from normal table is -0.674
P( x-/s.d < x - 15/3) = 0.25
That is, ( x - 15/3 ) = -0.67
--> x = -0.67 * 3 + 15 = 12.99
Q2 = P ( Z < x ) = 0.5
Value of z to the cumulative probability of 0.5 from normal table is 0
P( x-/s.d < x - 15/3 ) = 0.5
That is, ( x - 15/3 ) = 0
--> x = 0 * 3 + 15 = 15
Q3 = P ( Z < x ) = 0.75
Value of z to the cumulative probability of 0.75 from normal table is 0.674
P( x-/s.d < x - 15/3 ) = 0.75
That is, ( x - 15/3 ) = 0.67
--> x = 0.67 * 3 + 15 = 17.01   

b.
P ( Z < x ) = 0.90
Value of z to the cumulative probability of 0.90 from normal table is 1.30
P( x-/s.d < x - 15/3 ) = 0.90
That is, ( x - 15/3 ) = 1.30
--> x = 1.30 * 3 + 15 = 18.9   
c.
P ( Z > x ) = 0.65
Value of z to the cumulative probability of 0.65 from normal table is -0.3853
P( x-/ (s.d) > x - 15/3) = 0.65
That is, ( x - 15/3) = -0.3853
--> x = -0.3853 * 3+15 = 13.8441   
d.
P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.96
P( x-/s.d < x - 15/3 ) = 0.025
That is, ( x - 15/3 ) = -1.96
--> x = -1.96 * 3 + 15 = 9.12   
P ( Z > x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is 1.96
P( x-/ (s.d) > x - 15/3) = 0.025
That is, ( x - 15/3) = 1.96
--> x = 1.96 * 3+15 = 20.88   
under the corresponding normal curve into a middle area of 0.95 are [ 9.12, 20.88]

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