Question

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

Answer #1

Binomial distribution: P(X) = nCx p^{x}
q^{n-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 0.4^{6} x 0.6^{2}

= **0.0413**

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

= 0.0413 + 8x0.4^{7}x0.6 + 0.4^{8}

= 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)

= 8C3x0.4^{3}x0.6^{5} +
8C4x0.4^{4}x0.6^{4} +
8C5x0.4^{5}x0.6^{3}

= 0.2787 + 0.2322 + 0.1239

= **0.6348**

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