Question

5. Jockeys in the United States and England work very hard to keep their weight down. Many participate in weight-loss programs, carefully monitor their diet, and exercise regularly. The weight of a male jockey is approximately normal with mean 52 kg and standard deviation 1.2 kg. Suppose a male jockey is randomly selected. (a) What is the probability that the jockey weighs more than 53 kg? (b) What is the probability that the jockey weighs between 50 and 54 kg? (c) Find a value w such that 80% of all male jockeys weigh more than w.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 52

standard deviation = $\sigma$ = 1.2

(a)

P(x > 53) = 1 - P(x < 53)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (53 - 52) / 1.2]

= 1 - P(z < 0.83)

= 1 - 0.7967

= 0.2033

Probability = 0.2033

(b)

P(50 < x < 54) = P[(50 - 52)/ 1.2) < (x - $\mu$ ) /$\sigma$  < (54 - 52) / 1.2) ]

= P(-1.67 < z < 1.67)

= P(z < 1.67) - P(z < -1.67)

= 0.9525 - 0.0475

= 0.905

Probability = 0.905

(c)

Using standard normal table ,

P(Z > z) = 80%

1 - P(Z < z) = 0.8

P(Z < z) = 1 - 0.8

P(Z < -0.84) = 0.2

z = -0.84

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = -0.84 * 1.2 + 52 = 50.99

Answer = 50.99