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wariber [46]
3 years ago
14

An office manager has received a report from a consultant that includes a section on equipment replacement. The report indicates

that scanners have a service life that is normally distributed with a mean of 41 months and a standard deviation of 4 months. On the basis of this information, determine the percentage of scanners that can be expected to fail in the following time periods:
a. Before 38 months of service
b. Between 40 and 45 months of service
c. Within ± 2 months of the mean life
Business
1 answer:
goldenfox [79]3 years ago
5 0

Answer:

a) 22.663%

b) 44%

c) 38.3%

Explanation:

An office manager has received a report from a consultant that includes a section on equipment replacement. The report indicates that scanners have a service life that is normally distributed with a mean of 41 months and a standard deviation of 4 months. On the basis of this information, determine the percentage of scanners that can be expected to fail in the following time periods:

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score

μ is the population mean = 41 months

σ is the population standard deviation = 4 months

a. Before 38 months of service

Before in z score score means less than 38 months

Hence,

z = 38 - 41/4

z = -0.75

Probability value from Z-Table:

P(x<38) = 0.22663

Converting to percentage = 0.22663 × 100

= 22.663%

b. Between 40 and 45 months of service

For x = 40 months

z = 40 - 41/4

z = -0.2

Probability value from Z-Table:

P(x = 40) = 0.40129

For x = 45

z = 45 - 41/4

z = 1

Probability value from Z-Table:

P(x = 45) = 0.84134

Between 40 and 45 months of service

= 0.84134 - 0.40129

= 0.44005

Converting to Percentage

= 0.44005 × 100

= 44.005%

= 44%

c. Within ± 2 months of the mean life

+ 2 months = 41 months + 2 months

= 43 months

- 2 months = 41 months - 2 months

= 39 months

For x = 43

z = 43 - 41 /4

z = 0.5

P-value from Z-Table:

P(x = 43) = 0.69146

For x = 39

z = 39 - 41/4

z = -2/4

z = -0.5

Probability value from Z-Table:

P(x = 39) = 0.30854

Within ± 2 months of the mean life

= 0.69146 - 0.30854

= 0.38292

= 38.3%

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Turner, a successful executive, is negotiating a compensation plan with his potential employer. The employer has offered to pay
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Answer:

b. If the employer accepts Turner's counteroffer, Turner will recognize as gross income $55,000 per month [($480,000 + $180,000)/12].

Explanation:

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And, $180,000 bonus in 5 years at the age of 65

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6 0
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Managers should act in shareholders' interests because shareholders have ___________ priority in receiving their claims.A. TopB.
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Answer:

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Explanation:

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​A bond with a 12 percent quarterly coupon rate has a yield to maturity of 16 percent. The bond has a par value of $1,000 and ma
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Answer:

fair value is $761

Explanation:

Given data

bond value = $1000

rater r = 12 %

rate R = 16%

time = 20 year

to find out

a fair price

solution

we know compounding period in year is = 4

so time 20 x 4 = 80  

fair Price = \sum_{k=1}^{k=80} [(Quarterly Coupon) / (1 + R/400)^t] +bond value / (1 + R /400)^t

here

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so

fair Price = \sum_{k=1}^{k=80} [(30) / (1 + 16/400)^k] + 1000 / (1+16/400)^80

solve it we get

fair value is $761

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