Question

The actual tracking weight of a stereo cartridge that is set to track at 3 g on a particular changer can be regarded as a continuous rv X with the following pdf.

f(x) =

k 1 − (x − 3)2		2 ≤ x ≤ 4
0		otherwise

(a) Sketch the graph of f(x).

(b) Find the value of k.

(c) What is the probability that the actual tracking weight is greater than the prescribed weight?

(d) What is the probability that the actual weight is within 0.5 g of the prescribed weight? (Round your answer to four decimal places.)

(e) What is the probability that the actual weight differs from the prescribed weight by more than 0.6 g? (Round your answer to four decimal places.)

Answer 1

(a)

At x = 2.5, f(2.5) = 1- (2.5-3)² = 0.75   

We see that last graph (lower right) passes through (2.5, 0.75). The correct graph is last graph (lower right).

(b)

For a valid pdf,

k(1- (1 - 3)2) d.= 1

* [(x - (x – 3)3/3) = 1

=> k[(4 - (4-3)³/3 - 2 + (2 - 3)³/3] = 1

=> 4k/3 = 1

=> k = 3/4

(c)

The cumulative pdf is,

F(x) = |(3/4)(1 – (x – 3)2) dx = (3/4)[(x – (0 - 3)3/3),

= (3/4) [x - (x-3)³/3 - 7/3]

Probability that the actual tracking weight is greater than the prescribed weight = P(X > 3)

= 1 - P(X < 3)

= 1 - F(3)

= 1 - (3/4) [3 - (3-3)³/3 - 7/3]

= 0.5

(d)
Probability that the actual weight is within 0.5 g of the prescribed weight = P(2.5 < X < 3.5)

= F(3.5) - F(2.5)

= (3/4) [3.5 - (3.5-3)³/3 - 7/3] - (3/4) [2.5 - (2.5-3)³/3 - 7/3]

= 0.6875

(e)

Probability that the actual weight differs from the prescribed weight by more than 0.6 g = P(X < 3 - 0.6) + P(X > 3 + 0.6)

= P(X < 2.4) + P(X > 3.6)

= P(X < 2.4) + 1 - P(X < 3.6)

= F(2.4) + 1 - F(3.6)

= (3/4) [2.4 - (2.4-3)³/3 - 7/3] + 1 - (3/4) [3.6 - (3.6-3)³/3 - 7/3]

= 0.208