Question

In a factory, the average number of chairs manufactured per day is 60, and the standard deviation is 5. If a day is selected at random, find these probabilities. SHow your work:

a. at least 68 chairs will be manufactured.

b. at most 57 chairs will be manufactured.

c. between 54 and 73 chairs will be manufactured.

Answer #1

Solution :

Given that,

mean = = 60

standard deviation = =5

P(x > 68) = 1 - P(x< 68)

= 1 - P[(x -) / < (68-60) /5 ]

= 1 - P(z <1.6 )

Using z table

= 1 - 0.9452

probability= 0.0548

b.

P(x< 57)

=P[(x -) / < (57-60) /5 ]

= P(z <-0.6 )

Using z table

=0.2743

probability= 0.2743

c

P(54< x < 73) = P[(54-60) /5 < (x - ) / < (73-60) /5]

= P(-1.2 < Z <0.24 )

= P(Z < 0.24) - P(Z < -1.2)

Using z table

=0.5948 - 0.1151

probability= 0.4797

LeBron
Factory
Number
of
Workers
Number
of
Machines
Output
(chairs
produced
per hour)
Marginal
Product of
Labor
Cost of
Workers
Cost of
Machines
Total
Cost
?1
?2
?5
?2
?2
?10
?3
?2
?20
?4
?2
?35
?5
?2
?55
?6
?2
?70
?7
?2
?80
Refer to Table above.
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above.
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LeBron Chair Factory costs $12 per hour. The cost of each machine
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