Question

 

If a variable is Normal (? = 10, ? = 1.2)

a. Calculate the probability that X enters between 10 and 12.

b. Calculate the X corresponding to the 80% percentile

c. Calculate the probability that X is greater than 9

d. If a sample of 15 data is taken, calculate the probability that the average is between 9.2 and 10.

e. If you take a sample of 10 data, calculate the probability that the total is between 105 and 120.

 

Answer 1

a) P(10 < X < 12)

= P((10 - )/ < (X - )/ < (12 - )/)

= P((10 - 10)/1.2 < Z < (12 - 10)/12)

= P(0 < Z < 0.17)

= P(Z < 0.17) - P(Z < 0)

= 0.5675 - 0.5

= 0.0675

b) P(X < x) = 0.8

P((X - )/ < (x - )/) = 0.8

or, P(Z < (x - 10)/12) = 0.8

or, (x - 10)/12 = 0.84

or, x = 0.84 * 12 + 10

or, x = 20.08

c) P(X > 9)

= P((X - )/ > (9 - )/)

= P(Z > (9 - 10)/12)

= P(Z > -0.08)

= 1 - P(Z < -0.08)

= 1 - 0.4681

= 0.5319

d) P(9.2 < < 10)

= P((9.2 - )/() < ( - )/() < (10 - )/())

= P((9.2 - 10)/(1.2/sqrt(15)) < Z < (10 - 10)/(1.2/sqrt(15)))

= P(-2.58 < Z < 0)

= P(Z < 0) - P(Z < -2.58)

= 0.5 - 0.0049

= 0.4951

e) P(10.5 < < 12)

= P((10.5 - )/() < ( - )/() < (12 - )/())

= P((10.5 - 10)/(1.2/sqrt(10)) < Z < (12 - 10)/(1.2/sqrt(10)))

= P(1.32 < Z < 5.27)

= P(Z < 5.27) - P(Z < 1.32)

= 1 - 0.9066

= 0.0934