Question

Given a normal distribution with µ = 47 and σ = 6, what is the probability that:

X < 39 or X > 44

X is between 37 and 46

7% of the values are less than what X value.

Between what two X values (symmetrically distributed around the mean) are 70% of the values?

Answer 1

Part a)
X ~ N ( µ = 47 , σ = 6 )
P ( 39 < X < 44 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 39 - 47 ) / 6
Z = -1.3333
Z = ( 44 - 47 ) / 6
Z = -0.5
P ( -1.33 < Z < -0.5 )
P ( 39 < X < 44 ) = P ( Z < -0.5 ) - P ( Z < -1.33 )
P ( 39 < X < 44 ) = 0.3085 - 0.0912
P ( 39 < X < 44 ) = 0.2173

Required probability = 1 - 0.2173 = 0.7827

Part b)
X ~ N ( µ = 47 , σ = 6 )
P ( 37 < X < 46 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 37 - 47 ) / 6
Z = -1.6667
Z = ( 46 - 47 ) / 6
Z = -0.1667
P ( -1.67 < Z < -0.17 )
P ( 37 < X < 46 ) = P ( Z < -0.17 ) - P ( Z < -1.67 )
P ( 37 < X < 46 ) = 0.4338 - 0.0478
P ( 37 < X < 46 ) = 0.3860

Part c)
X ~ N ( µ = 47 , σ = 6 )
P ( X < x ) = 7% = 0.07
To find the value of x
Looking for the probability 0.07 in standard normal table to calculate Z score = -1.4758
Z = ( X - µ ) / σ
-1.4758 = ( X - 47 ) / 6
X = 38.1452
P ( X < 38.1452 ) = 0.07

Part d)
X ~ N ( µ = 47 , σ = 6 )
P ( a < X < b ) = 0.7
Dividing the area 0.7 in two parts we get 0.7/2 = 0.35
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.35
Area above the mean is b = 0.5 + 0.35
Looking for the probability 0.15 in standard normal table to calculate Z score = -1.0364
Looking for the probability 0.85 in standard normal table to calculate Z score = 1.0364
Z = ( X - µ ) / σ
-1.0364 = ( X - 47 ) / 6
a = 40.7816
1.0364 = ( X - 47 ) / 6
b = 53.2184
P ( 40.7816 < X < 53.2184 ) = 0.7