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1
O N L I N E T U T O R I A L
Statistical Tools for Managers
DISCUSSION QUESTIONS
1. A probability distribution is a statement of a probability function,
which assigns all the probabilities associated with a random variable.
3. The variance is a measure of the dispersion of the distribution.
4. Examples of business processes that might be described by a
normal distribution could include sales of a product, project com-
Expected value = 5.45
ii
X = X p =
2
Variance = ( ) 4.06
ii
X X p − =
T1.3 Here: x = 280,
= 250,
= 25. Given:
360 ONLINE TUTORIAL 1 STA TIS T I CA L TOO L S FO R MA NA G E RS
ONLINE TUTORIAL 1 STA T I STI CA L TO O LS FO R MA N A G E R S 361
Because the table in Appendix I handles only positive z
Thus, we need to find the area to the left of 1
, i.e.,
P(z 1) Because the table in Appendix I handles only
positive values of z, we need to determine the area to
the right of z = 1. This area is obtained by finding the
probability:
( 1) 1 ( 1)
1 0.8413
0.1587
P z P z = −
=−
=
(Where 0.8413 = P(z 1) is obtained from the table in
Appendix I.)
Thus, the probability that the shaft has a diameter
less than or equal to 0.45 inches is 0.1587.
362 ONLINE TUTORIAL 1 STA TIS T I CA L TOO L S FO R MA NA G E RS
1
2
1
1
2
2
470
460
450
25
470 450 20
25 25
0.8
460 450 10
25 25
0.4
x
x
x
z
z
z
z
z
=
=
=
=
−
=
−
==
=
−
==
=
Therefore:
= − (460 470) ( 0.8) ( 0.4)P x P z P z
From the table in Appendix I:
=
=
( 0.8) 0.7881
( 0.4) 0.6554
Pz
Pz
and
= − =(460 470) 0.7881 0.6554 0.1327Px
T1.8 (a) The probability that sales will be greater than 5,500
From the table in Appendix I, P(z 1.6) = 0.9452.
( 5,500) ( 1.6)
1 ( 1.6)
1 0.9452
0.0548
P x P z
Pz
=
= −
=−
=
Therefore, the probability of selling at least 5,500
oranges is 0.0548.
(b) The probability that sales will be greater than 4,500
oranges is found as follows:
−
=
−−
==
=−
− =
4500 4700 200
500 500
0.4
( 0.4) ( 0.4)
x
z
P z P z
ONLINE TUTORIAL 1 STA T I STI CA L TO O LS FO R MA N A G E R S 363
P(z 0.4) = 0.6554.
Therefore, the probability that sales will be less than
4,900 oranges is 0.6554. (Note that the answer to this
question is the same as to question T1.8b—the normal
distribution is symmetrical.)
(d) The probability that sales will be less than 4,300 oranges
= −
1 ( 0.8)
Pz
From the table in Appendix I, P(z 0.8) = 0.7881.
Therefore:
( 0.8) 1 0.7881
0.2119
Pz − = −
=
Thus, the probability of selling 4,300 or fewer oranges
T1.9 The probability that sales will be 81,000 or fewer units is
found as follows:
81,000
87,000
4,000
81,000 87,000
4,000
6,000
1.5
x
x
z
=
=
=
−
=
−
=
−
=
=−
− = = − ( 1.5) ( 1.5) 1 ( 1.5)P z P z P z
