MTH302 Practice Questions Lectures 41-45 Updated

MCQS

The standard normal distribution is:

a) skewed to the right. b) also called the t distribution. c) an approximation of the binomial distribution. d) defined by a mean equal to 0 and a standard deviation equal to 1.**

X is a random normal variable, with mean μ and variance σ2. The “standardized form” of X is X Z . What are the mean and variance, respectively, of Z?

a) 2, 0 b) 1, 0 c) 0, 1** d) 2, 1

Let X∼N (4, 22). Which of the following is a standard normal variable?

a) X Z

b) X Z

c) X Z

d) X Z

The normal distribution is a _____ distribution.

a) Symmetrical** b) Positively Skewed c) Negatively Skewed d) Uniform

P(0≤Z≤2) = P(0≤Z≤2)

= 0.4772.  = 36. Find the probability that it will be smaller than 57.

P(X<57) = (Z 2) X P

P

= P(-∞<Z<0) + P(0<Z<2)

= 0.5 + 0.4772 = 0.9772.

Suppose that weights of 2000 male students are normally distributed with mean 155 pounds and standard deviation 20 pounds. Find the probability that:

(i) The students weight lie between 120 and 130 pounds. (ii) The students weight less than or equal to 100 pounds.

Solution:

P(120<X<130) = ( 1.75 Z 1.25) X P

P = P(Z<-1.75) - P(Z<-1.25)

= 0.4599 – 0.3944 = 0.0655

(ii)

P(X<100) = (Z 2.75) X P

P

= P(-∞<Z<0) – P(-2.75<Z<0) = 0.5 – 0.4970 = 0.003.

A soft drink machine is regulated so that it discharges an average of 200 milliliters per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 milliliters,

a) What fraction of the cups will contain more than 220 milliliters?

1. What is the probability that a cup contains between 180 and 200 milliliters?

(Probability of z equal to 1.33 is 0.4082)

Solution:

P(X>220) = (Z 1.33) X P

P