Question

# From past experience, a restaurant owner has determined that 40% of those who visit his restaurant...

From past experience, a restaurant owner has determined that 40% of those who visit his restaurant will spend at least \$75. If 8 people will visit his restaurant tomorrow, find the probability that:
a) 6 will spend at least 75.
b) at least 6 will spend at least \$75
c) between 3 and 5, inclusive, will spend at least \$75

Binomial distribution: P(X) = nCx px qn-x

Sample size, n = 8

P(a person will spend at least \$75), p = 0.4

q = 1 - p = 0.6

a) P(6 will spend at least \$75) = P(X = 6)

= 8C6 x 0.46 x 0.62

= 0.0413

b) P(at least 6 will spend at least \$75) = P(6) + P(7) + P(8)

= 0.0413 + 8x0.47x0.6 + 0.48

= 0.0413 + 0.0079 + 0.0007

= 0.0499

c) P(between 3 and 5, inclusive, will spend at least \$75) = P(3) + P(4) + P(5)

= 8C3x0.43x0.65 + 8C4x0.44x0.64 + 8C5x0.45x0.63

= 0.2787 + 0.2322 + 0.1239

= 0.6348

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