Trevor is interested in purchasing the local hardware/sporting goods store in th
ID: 3290454 • Letter: T
Question
Trevor is interested in purchasing the local hardware/sporting goods store in the small town of Dove Creek, Montana. After examining accounting records for the past several years, he found that the store has been grossing over $850 per day about 65% of the business days it is open. Estimate the probability that the store will gross over $850 for the following. (Round your answers to three decimal places.) (a) at least 3 out of 5 business days (b) at least 6 out of 10 business days (c) fewer than 5 out of 10 business days (d) fewer than 6 out of the next 20 business days If the outcome described in part (d) actually occurred, might it shake your confidence in the statement p = 0.65? Might it make you suspect that p is less than 0.65? Explain. Yes. This is unlikely to happen if the true value of P is 0.65. Yes. This is likely to happen in the true value of P is 0.65. No. This is unlikely to happen if the true value of P is 0.65. NO. This is likely to happen if the true value of P is 0.65. (e) more than 17 out of the next 20 business days If the outcome described in part (e) actually occurred, might you suspect that P is greeter than 0.65? Explain. Yes. This is unlikely to happen if the true value of P is 0.65. Yes. This is likely to happen if the true value of p is 0.65. No. This is unlikely to happen if the true value of p is 0.65. No. This is likely to happen if the true value of P is 0.65.Explanation / Answer
This is the case of Binomial Distribution:
p= 0.65
p (x) = C (n,x) * p^x*(1-p)^(n-x)
Question a)
x
Formula
P (x)
3
C(5,3) * 0.65^3 * 0.35^2
0.336
4
C(5,4) * 0.65^4 * 0.35^1
0.312
5
C(5,5) * 0.65^5 * 0.35^0
0.116
P (x>= 3 ) = p (x = 3) + P ( x = 4) + P (x=5)
= 0.336 + 0.312 + 0.116
= 0.764
Answer: 0.764
Question b)
x
Formula
P (x)
6
C(10,6) * 0.65^6 * 0.35^4
0.238
7
C(10,7) * 0.65^7 * 0.35^3
0.252
8
C(10,8) * 0.65^8 * 0.35^2
0.176
9
C(10,9) * 0.65^9 * 0.35^1
0.072
10
C(10,10) * 0.65^10 * 0.35^0
0.013
P (x>=6 ) = P ( x= 6 ) + P ( x = 7 ) + P ( x = 8 ) + P ( x = 9 ) + P ( x = 10)
= 0.238 + 0.252 + 0.176 + 0.072 + 0.013
= 0.751
Answer: 0.751
Question c)
x
Formula
P (x)
0
C(10,0) * 0.65^0 * 0.35^10
0.000
1
C(10,1) * 0.65^1 * 0.35^9
0.001
2
C(10,2) * 0.65^2 * 0.35^8
0.004
3
C(10,3) * 0.65^3 * 0.35^7
0.021
4
C(10,4) * 0.65^4 * 0.35^6
0.069
P (x < 5)
= p (x= 0) + p (x=1) + p (x= 2) + p (x=3) + p (x=4)
= 0.000 + 0.001 + 0.004 + 0.021 + 0.069
= 0.095
Answer: 0.095
Question d)
x
Formula
P (x)
0
C(20,0) * 0.65^0 * 0.35^20
0.000
1
C(20,1) * 0.65^1 * 0.35^19
0.000
2
C(20,2) * 0.65^2 * 0.35^18
0.000
3
C(20,3) * 0.65^3 * 0.35^17
0.000
4
C(20,4) * 0.65^4 * 0.35^16
0.000
5
C(20,5) * 0.65^5 * 0.35^15
0.000
P (x< 6) = P (x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)
= 0.000 +0.000 + 0.000 + 0.000 +0.000 + 0.000
= 0.000
Answer: 0.000
Yes. This is unlikely to happen if the true value of p is 0.65
Question e)
x
Formula
P (x)
18
C(20,18) * 0.65^18 * 0.35^2
0.010
19
C(20,19) * 0.65^19 * 0.35^1
0.002
20
C(20,20) * 0.65^20 * 0.35^0
0.000
P ( x > 17) = P ( x = 18 ) + P ( x = 19 ) + P ( x = 20 )
= 0.01 + 0.002 + 0.000
= 0.012
Answer: 0.012
Yes. This is unlikely to happen if the true value of p is 0.65
x
Formula
P (x)
3
C(5,3) * 0.65^3 * 0.35^2
0.336
4
C(5,4) * 0.65^4 * 0.35^1
0.312
5
C(5,5) * 0.65^5 * 0.35^0
0.116
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