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Andrej [43]
3 years ago
15

John is interested in purchasing a multi-office building containing five offices. The current owner provides the following proba

bility distribution indicating the probability that the given number of offices will be leased each year. Number of Lease Offices 0 1 2 3 4 5 Probability 5/18 1/4 1/9 1/18 2/9 1/12 If each yearly lease is $12,000, how much could John expect to collect in yearly leases for the whole building in a given year?(in dollars)
a) E(X) = $23,353.33
b) E(X) = $23,333.33
c) E(X) = $23,273.33
d) E(X) = $23,263.33
e) E(X) = $23,423.33
f) None of the above.
Mathematics
1 answer:
Elenna [48]3 years ago
3 0

Answer:

Option B.

Step-by-step explanation:

The given table is:

Number of Lease Offices :  0           1         2       3        4         5

Probability                         : 5/18     1/4     1/9    1/18     2/9      1/12

The expected probability is

Expected probability = \sum_{i=0}^5 x_{i}p(x_i)

Expected probability = 0p(0)+1P(1)+2P(2)+3P(3)+4P(4)+5P(5)

Expected probability = 0\cdot (\frac{5}{18})+1\cdot (\frac{1}{4})+2\cdot (\frac{1}{9})+3\cdot (\frac{1}{18})+4\cdot (\frac{2}{9})+5\cdot (\frac{1}{12})=\frac{35}{18}

It is given that the yearly lease = $12,000.

The yearly leases for the whole building in a given year is

Yearly leases = \frac{35}{18}\times 12000=23333.3333333\approx 23333.33

Therefore, the correct option is B.

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A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 19 randomly selected pens yield
Romashka [77]

Given Information:  

Probability of shipment accepted = p = 5%

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Required Information:  

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Answer:

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Step-by-step explanation:

The given problem can be solved using Bernoulli distribution  which is given by

P(n, x) = nCx pˣqⁿ⁻ˣ  

The probability of no more than 2 defective pens means

P( x ≤ 2) = Probability of 0 defective pen + Probability of 1 defective pen + Probability of 2 defective pens

P( x ≤ 2) = P(0) + P(1) + P(2)

For P(0) we have p = 0.05, q = 0.95, n = 19 and x = 0

P(0) = 19C0(0.05)⁰(0.95)¹⁹

P(0) = (1)(1)(0.377)

P(0) = 0.377

For P(1) we have p = 0.05, q = 0.95, n = 19 and x = 1

P(1) = 19C1(0.05)¹(0.95)¹⁸

P(1) = (19)(0.05)(0.397)

P(1) = 0.377

For P(2) we have p = 0.05, q = 0.95, n = 19 and x = 2

P(2) = 19C2(0.05)²(0.95)¹⁷

P(2) = (171)(0.0025)(0.418)

P(2) = 0.179

Therefore, the required probability is

P( x ≤ 2) = P(0) + P(1) + P(2)

P( x ≤ 2) = 0.377 + 0.377 + 0.179

P( x ≤ 2) = 0.933

P( x ≤ 2) = 93.3%

Therefore, the probability that this shipment is accepted with no more than 2 defective pens is 0.933.

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