Question

The lengths of a professor's classes has a continuous uniform distribution between 50.0 min and 52.0 min. If one such class is randomly selected, find the probability that the class length is between 51.1 and 51.5 min. Round the answer to 3 decimal places. P(51.1 < X < 51.5) =

For a standard normal distribution, find: P(z > 2.04)

For a standard normal distribution, find: P(-1.04 < z < 2.04)

For a standard normal distribution, find: P(z > c) = 0.6332 Find c.

Answer 1

The probability density function and cumulative distribution function for a continuous uniform distribution : X on the interval [a,b] are

= 0 otherwise

X : lengths of a professor's classes

X follows uniform distribution between 50.0 min and 52.0 min.

Probability density function of X

= 0 otherwise

Cumulative Distribution function : P(X<x)

Probability that the class length is between 51.1 and 51.5 min = P(51.1 < X < 51.5) =P(X<51.5) - P(X<51.1)

P(51.1 < X < 51.5) =P(X<51.5) - P(X<51.1) = 0.75-5.5 = 0.20

Probability that the class length is between 51.1 and 51.5 min = P(51.1 < X < 51.5) = 0.200

For a standard normal distribution, find: P(z > 2.04)

P(z > 2.04) = 1-P(z2.04)

From standard normal tables, P(z 2.04) = 0.9793

P(z > 2.04) = 1-P(z2.04) = 1 -0.9793= 0.0207

P(z > 2.04) = 0.0207

For a standard normal distribution, find: P(-1.04 < z < 2.04)

P(-1.04 < z < 2.04) = P(z<2.04) - P(z<-1.04)

From Standard normal tables , P(z 2.04) = 0.9793 ; P(z<-1.04)=0.1492

P(-1.04 < z < 2.04) = P(z<2.04) - P(z<-1.04) = 0.9793 - 0.1492= 0.8301

P(-1.04 < z < 2.04) = 0.8301

For a standard normal distribution, find: P(z > c) = 0.6332 Find c.

P(zc) = 1-P(z > c) = 1-0.6332= 0.3668

From standard normal tables, c = -0.34 such that P(z<c) = 0.3669(nearest to 0.3668)

c = - 0.34