Question

1. The weight of Los Angeles Angels players is evenly distributed with a mean of 203 pounds and a standard deviation of 18 pounds (round to the nearest second decimal place)

a. What is the probability of a player weighing more than 220 pounds?

b. What is the probability of a player weighing less than 190 pounds?

c. What percent of players weigh between 188 and 218 pounds?

Answer 1

Solution :

Given that ,

mean = = 203

standard deviation = = 18

(a)

P(x > 220) = 1 - P(x < 220)

= 1 - P((x - ) / < (220 - 203) / 18)

= 1 - P(z < 0.94)

= 1 - 0.8264   

= 0.1736

Probability = 0.1736

(b)

P(x < 190) = P((x - ) / < (190 - 203) / 18)

= P(z < -0.72)

= 0.2358

Probability = 0.2358

(c)

P(188 < x < 218) = P((188 - 203)/ 18) < (x - ) /  < (218 - 203) / 18) )

= P(-0.83 < z < 0.83)

= P(z < 0.83) - P(z < -0.83)

= 0.7967 - 0.2033

= 0.5934

Answer = 59.34%