Given 𝑿 is a normally distributed random variable with a mean of 155 and a standard deviation of 11.6, determine

Question

Given 𝑿 is a normally distributed random variable with a mean of 155 and a standard deviation of 11.6, determine

A) 𝑷(𝑿>170) B) 𝑷(𝟏𝟓𝟐<𝑋<170)

Answer 1

X ~ N(155,11.6)

a) P(X>170) = 1 - P(X<170)

Calculate the standard normal variant Z = (170 - 155)/11.6 = 1.2931

P(X<170) = P(Z < 1.2931) = Φ(Z=1.2931) = 0.9020

Hence:

P(X>170) = 1 - P(X<170) = 1 - 0.9020 = 0.098

b)𝑷(𝟏𝟓𝟐<𝑋<170) = P(X<170) - P(X<152)

Z1 = (170-155)/11.6 = 1.2931 --> Φ(Z=1.2931) = 0.9020

Similarly

Z2 = (152-155)/11.6 = -0.2586 --> Φ(Z=-0.2586) = 0.3980

Therefore 𝑷(𝟏𝟓𝟐<𝑋<170) = P(X<170) - P(X<152) = 0.9020 - 0.3980 = 0.504

Matlab codes

a) normcdf(170,155,11.6)

ans =

0.9020

b) normcdf(170,155,11.6)-normcdf(152,155,11.6)

ans =

0.5040