Question

1. What is the mean and standard deviation associated with the standard normal distribution?

2. Use the standard normal distribution table to calculate the following probabilities

A) P (z ≤ -1.64)

B) P (z ≤ 2.11)

C) P (z ≥ 0.73)

D) P (-1.31 ≤ z ≤ -0.32)

3. Use the standard normal distribution table to find the z-value for each of the following

A) P (z ≤ ?) = 0.9495

B) P (z ≤ ?) = 0.2810

C) P (z ≥ ?) = 0.9968

D) P ( -1.13 ≤ z ≤ ?) = 0.8299

4. A local grocery store owner wants to learn more about how many apples patrons buy from his store week to week, and he has asked for your help calculating some probabilities. He tells you that he believes the data to be normally distributed, and that the average amount of apples bought each week is 678.32 lbs. with a standard deviation of 53.98 lbs.

A) What is the probability that the store sells more than 750 lbs. of apples in a week?

B) What is the probability that the store sells less than 500 lbs. of apples in a week?

C) What is the probability that the store sells between 600 and 700 lbs. of apples?

D) What is the probability that the store sells exactly 678.32 lbs. of apples?

Answer 1

Question 1

Standard normal distribution has mean µ = 0 and standard deviation σ = 1

Question 2

a) P ( Z <= -1.64 ) = 0.0505

b)   P ( Z <= 2.11 ) = 0.9826

c)

P ( Z >= 0.73 ) = 1 - P ( Z < 0.73 )
P ( Z >= 0.73 ) = 1 - 0.7673
P ( Z >= 0.73 ) = 0.2327

d)

P ( -1.31 < Z < -0.32 ) = P ( Z < -0.32 ) - P ( Z < -1.31 )
P ( -1.31 < Z < -0.32 ) = 0.3745 - 0.0951
P ( -1.31 < Z < -0.32 ) = 0.2794

Question 3

X ~ N ( µ = 0 , σ = 1 )
P ( X < x ) = 94.95% = 0.9495
To find the value of x
Looking for the probability 0.9495 in standard normal table to calculate Z score = 1.64
P ( Z <= 1.64 ) = 0.9495

b)

X ~ N ( µ = 0 , σ = 1 )
P ( X < x ) = 28.1% = 0.281
To find the value of x
Looking for the probability 0.281 in standard normal table to calculate Z score = -0.58
P ( Z <= -0.58 ) = 0.2810

c)

X ~ N ( µ = 0 , σ = 1 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.9968 = 0.0032
To find the value of x
Looking for the probability 0.0032 in standard normal table to calculate Z score = -2.73
P ( Z >= -2.73 ) = 0.9968

d)

X ~ N ( µ = 0 , σ = 1 )
P ( a < X < b ) = 0.8299
Standardizing the value
P ( a < Z < b ) = P ( Z < b ) - P ( Z < a )
Need to find the value of b
0.8299 = P(Z < -1.13) - P(Z < a)
P ( Z < a ) = 0.8299 + 0.1292
P ( Z < a ) = 0.9591
Looking for the probability 0.9591 in standard normal table to calculate Z score = 1.74
P ( -1.13 < X < 1.74 ) = 0.8299

Question 4

a)

X ~ N ( µ = 678.32 , σ = 53.98 )
P ( X > 750 ) = 1 - P ( X < 750 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 750 - 678.32 ) / 53.98
Z = 1.3279
P ( ( X - µ ) / σ ) > ( 750 - 678.32 ) / 53.98 )
P ( Z > 1.3279 )
P ( X > 750 ) = 1 - P ( Z < 1.3279 )
P ( X > 750 ) = 1 - 0.9079
P ( X > 750 ) = 0.0921

b)

X ~ N ( µ = 678.32 , σ = 53.98 )
P ( X < 500 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 500 - 678.32 ) / 53.98
Z = -3.3034
P ( ( X - µ ) / σ ) < ( 500 - 678.32 ) / 53.98 )
P ( X < 500 ) = P ( Z < -3.3034 )
P ( X < 500 ) = 0.0005

c)

X ~ N ( µ = 678.32 , σ = 53.98 )
P ( 600 < X < 700 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 600 - 678.32 ) / 53.98
Z = -1.4509
Z = ( 700 - 678.32 ) / 53.98
Z = 0.4016
P ( -1.45 < Z < 0.4 )
P ( 600 < X < 700 ) = P ( Z < 0.4 ) - P ( Z < -1.45 )
P ( 600 < X < 700 ) = 0.656 - 0.0734
P ( 600 < X < 700 ) = 0.5826

d)

X ~ N ( µ = 678.32 , σ = 53.98 )
P ( 677.82 < X < 678.82 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 677.82 - 678.32 ) / 53.98
Z = -0.0093
Z = ( 678.82 - 678.32 ) / 53.98
Z = 0.0093
P ( -0.01 < Z < 0.01 )
P ( 677.82 < X < 678.82 ) = P ( Z < 0.01 ) - P ( Z < -0.01 )
P ( 677.82 < X < 678.82 ) = 0.5037 - 0.4963
P ( 677.82 < X < 678.82 ) = 0.0074