Let X = the number of drivers that wear their seat belt while driving. In Evansv
ID: 3241376 • Letter: L
Question
Let X = the number of drivers that wear their seat belt while driving. In Evansville, it has been found that in a survey of 500 drivers only 40% of the drivers on average wear their seatbelts regularly. If X is NORMALLY distributed with standard deviation of 30 drivers, what is the probability that between 170 and 220 (inclusive) of the drivers in a random sample regularly wear a seatbelt? If X is UNIFORMLY distributed with an upper limit of 260 drivers and a lower limit of 145 drivers, what is the probability that between 170 and 220 (inclusive) of the drivers in a random sample would wear their seatbelt?Explanation / Answer
a)
X ~Bin(n,p) = Bin(500,0.4)
E(x) = np = 500*0.4 = 200
p( 170 <= x <= 220) = p(x<= 220) - p(x <= 170)
= p(x<220.5) - p(x<170.5) (using continuty correction , p(X<=n) = p(X<n+0.5)
= p(z< 220.5 - 200/30) - p(z<170.5 - 200/30)
= p( z< 0.6833) - p( z< -0.9833)
= p(z<0.6833) - (1-p(z<0.9833))
= 0.75279 - 0.16273
= 0.5901
b)
X~ Uniform( 260,145)
E(X) = 1/2 (a+b) = 202.5
var(x) = 1/12 (b-a)2
= 33.2
p(170<=x<=220) = p(x<=220) - p(x<=170)
= p(z< 220-202.5 / 33.2) - p(z<170-202.5/33.2)
= p(z<0.5271)-p(z< -0.9789)
= p(z<0.5271) - (1-p(z<0.9789))
= 0.70094 - 0.16381
= 0.53713
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