In: Statistics and Probability
Trevor is interested in purchasing the local hardware/electronic goods store in a small town in South Ohio. After examining accounting records for the past several years, he found that the store has been grossing over $850 per day about 60% of the business days it is open. Estimate the probability that the store will gross over $850
0.68256
Answer:
Given,
p = 0.6
q = 1 - p
= 1 - 0.6
= 0.4
Binomial distribution P(X = r) = nCr*p^r*q^(n-r)
nCr = n!/(n-r)!*r!
a)
P(at least 6) = P(6) + P(7) + P(8) + P(9) + P(10)
= (10C6) * (0.6)^6 * (0.4)^4 + (10C7) * (0.6)^7 * (0.4)^3 + (10C8) * (0.6)^8 * (0.4)^2 + (10C9) * (0.6)^9 * (0.4)^1 + (10C10) * (0.6)^10 * (0.4)^
On solving we get
= 0.25 + 0.214 + 0.12 + 0.0403 + 0.006
= 0.63
b)
P(at most 3) = P(0) + P(1) + P(2) + P(3)
= (10C0) * .6^1 * 0.4^10 + (10C1) * 0.6^1 * .4^9 + (10C2) * 0.6^2 * 0.4 ^ 8 + (10C3) * 0.6 ^ 3 * 0.4^7
On solving we get
= 0.058
c)
P(fewer than 7) = P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6)
= (20C0) * 0.6^0 * 0.4^20 + (20C1) * 0.6^1 * 0.4^19 + (20C2) * 0.6^2 * 0.4^18 + (20C3) * 0.6^3 * 0.4^17 +(20C4) * 0.6^4 * 0.4^16 + (20C5) * 0.6^5 * 0.4^15 + (20C6) * 0.6^6 * 0.4^14
On solving we get
= 0.00653
d)
P(more than 16) = P(17) + P(18) + P(19) + P(20)
= (20C17) * 0.6^17 * .4^3 + (20C18) * 0.6^18* .4^2 + (20C19) * 0.6^19 * .4^1 + (20C20) * 0.6^20 * 0.4^0
On solving we get
= 0.016