Question

If a variable is Normal (µ = 10, σ = 1.2)
to. Find the probability that X is between 10 and 12. (10 points)
b. Calculate the X corresponding to the 80% percentile (10 points)
c. Find the probability that X is greater than 9 (10 points)
d. If a sample of 15 data is taken, calculate the probability that the average is between 9.2 and 10. (10 points)
and. Calculate the 90% percentile of the average of X if the sample is 15 data. (10 points)
F. If you sample 10 data, find the probability that the Total is between 105 and 120. (10 points)

Answer 1

a)

Given :- = 10 , = 1.2 )
We convet this to Standard Normal as
P(X < x) = P( Z < ( X - ) / )
P ( 10 < X < 12 ) = P ( Z < ( 12 - 10 ) / 1.2 ) - P ( Z < ( 10 - 10 ) / 1.2 )
= P ( Z < 1.67) - P ( Z < 0 )
= 0.9525 - 0.5
= 0.4525

b)

X ~ N ( µ = 10 , σ = 1.2 )
P ( X < x ) = 80% = 0.8
To find the value of x
Looking for the probability 0.8 in standard normal table to calculate critical value Z = 0.8416
Z = ( X - µ ) / σ
0.8416 = ( X - 10 ) / 1.2
X = 11.01

c)

P ( X > 9 ) = P(Z > (9 - 10 ) / 1.2 )
= P ( Z > -0.83 )
= 1 - P ( Z < -0.83 )
= 1 - 0.2033
= 0.7967

d)

Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 9.2 - 10 ) / ( 1.2 / √(15))
Z = -2.58
Z = ( 10 - 10 ) / ( 1.2 / √(15))
Z = 0
P ( 9.2 < X̅ < 10 ) = P ( Z < 0 ) - P ( Z < -2.58 )
P ( 9.2 < X̅ < 10 ) = 0.5 - 0.0049
P ( 9.2 < X̅ < 10 ) = 0.4951

e)

X ~ N ( µ = 10 , σ = 1.2 )
P ( X < x ) = 90% = 0.9
To find the value of x
Looking for the probability 0.9 in standard normal table to calculate critical value Z = 1.2816
Z = ( X - µ ) / ( σ / √(n) )
1.2816 = ( X - 10 ) / ( 1.2/√(15) )
X = 10.40

f)

X ~ N ( n , n 2 ) = N( 10 * 10 , 10 * 1.22 ) = N ( 100 , 3.794733 )

So,

P(105 < X < 120) =  P ( Z < ( 120 - 100 ) / 3.794733 ) - P ( Z < ( 105 - 100 ) / 3.794733 )
= P ( Z < 5.27) - P ( Z < 1.32 )
= 1 - 0.9066
= 0.0934