Question

4. The Federal Reserve reports that the mean lifespan of a five dollar bill is 4.9 years. Let’s suppose that the standard deviation is 1.9 years and that the distribution of lifespans is normal (not unreasonable!) Find: (a) the probability that a $5 bill will last more than 4 years. (b) the probability that a $5 bill will last between 3 and 5 years. (c) the 90th percentile for the lifespan of these bills (a time such that 90% of bills last less than that time). (d ) the probability that a random sample of 33 bills has a mean lifespan of more than 4.5 years.

Answer 1

Part a)

P ( X > 4 ) = 1 - P ( X < 4 )
Standardizing the value

Z = ( 4 - 4.9 ) / 1.9
Z = -0.47

P ( Z > -0.47 )
P ( X > 4 ) = 1 - P ( Z < -0.47 )
P ( X > 4 ) = 1 - 0.3192
P ( X > 4 ) = 0.6808

Part b)

P ( 3 < X < 5 )
Standardizing the value

Z = ( 3 - 4.9 ) / 1.9
Z = -1
Z = ( 5 - 4.9 ) / 1.9
Z = 0.05
P ( -1 < Z < 0.05 )
P ( 3 < X < 5 ) = P ( Z < 0.05 ) - P ( Z < -1 )
P ( 3 < X < 5 ) = 0.521 - 0.1587
P ( 3 < X < 5 ) = 0.3623

Part c)

P ( X < ? ) = 90% = 0.9
Looking for the probability 0.9 in standard normal table to calculate critical value Z = 1.28

1.28 = ( X - 4.9 ) / 1.9
X = 7.332 7.3
P ( X < 7.3 ) = 0.9

Part d)

P ( X > 4.5 ) = 1 - P ( X < 4.5 )
Standardizing the value


Z = -1.21

P ( Z > -1.21 )
P ( X > 4.5 ) = 1 - P ( Z < -1.21 )
P ( X > 4.5 ) = 1 - 0.1133
P ( X > 4.5 ) = 0.8867