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 97th percentile for the lifespan of these bills (a time such that 97% of bills last less than that time). (d ) the probability that a random sample of 37 bills has a mean lifespan of more than 4.5 years.

Answer 1

4)

Solution :

Given that ,

mean = = 4.9

standard deviation = = 1.9

(a)

P(x > 4) = 1 - P(x < 4)

= 1 - P((x - ) / < (4 - 4.9) / 1.9)

= 1 - P(z < -0.4737)

= 1 - 0.3179

= 0.6821

P(x > 4) = 0.6821

Probability = 0.6821

(b)

P(3 < x < 5) = P((3 - 4.9)/ 1.9) < (x - ) / < (5 - 4.9) / 1.9) )

= P(-1 < z < 0.0526)

= P(z < 0.0526) - P(z < -1)

= 0.521 - 0.1587

= 0.3623

Probability = 0.3623

(c)

P(Z < z) = 97%

P(Z < 1.88) = 0.97

z = 1.88

Using z-score formula,

x = z * +

x = 1.88 * 1.9 + 4.9 = 8.472

97th percentile = 8.472

(d)

n = 37

= 4.9 and

= / n = 1.9 / 37 = 0.3126

P( > 4.5) = 1 - P( < 4.5)

= 1 - P(( - ) / < (4.5 - 4.9) / 0.3126)

= 1 - P(z < -1.28)

= 1 - 0.1003

= 0.8997

P( >4.5) = 0.8997

Probability = 0.8997