5 Suppose only 40 of all drivers in Florida regularly wear a

5. Suppose only 40% of all drivers in Florida regularly wear a seatbelt. A random sample of 500 drivers is selected. What is the probability that a. Between 170 and 220 (inclusive) of the drivers in the sample regularly wear a seatbelt? b. Fewer than 175 of those in the sample regularly wear a seatbelt?

Solution

mean=n*p=500*0.4 =200

standard deviation =sqrt(n*p*(1-p)) =sqrt(500*0.4*0.6) =10.95445

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(a)P(170<X<220) = P((170-200)/10.95445<(X-mean)/s <(220-200)/10.95445)

=P(-2.74<Z<1.83) =0.9633 (from standard normal table)

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(b)P(X<175) = P(Z<(175-200)/10.95445)

=P(Z<-2.28) =0.0113(from standard normal table)