Laundromat Token
Probabilities – Normal Distribution?
Circular metal tokens are used to operate a washing machine in a laundromat. The diameters of the tokens are normally distributed, and only tokens with diameters between 1.94 and 2.06 will operate the machine. The mean = 2 and the standard deviation = 0.0305.
Find the probability that at most one token out of a randomly selected sample of 20 will not operate the machine.
Probability that a token will not operate the machine
= 1- P(1.94 < X < 2.06)
= 1 - P((1.94 - 2)/0.0305 < Z < (2.06 - 2)/0.0305 )
= 1 - P(-120/61 < Z < 120/61)
= 1 - 0.95084
= 0.04916
Binomial Distribution, n = 20
X~B(20, 0.04916)
P(X <= 1)
= P(X = 0) + P(X = 1)
= (20C0) (0.04916)^0 (1 - 0.04916)^20 + (20C1) (0.04916)^1 (1 - 0.04916)^(20 - 1)
= 0.742175948
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