In: Statistics and Probability
Assume the average weight of a player in the NFL is normally distributed with a population mean of 245 pounds with a population standard deviation of 46 pounds. Suppose we take a sample of 50 NFL players.
a. What is the probability that a randomly selected player will weigh over 300 pounds?
b. What is the probability that a randomly selected player will weigh under 180 pounds?
c. What is the probability that a randomly selected player will weigh between 190 and 320 pounds?
d. What is the probability that the average weight of players from the sample of 50 will be above 260?
e. What weight defines the lowest 34.09% of the distribution for an individual player?
f. What weight defines the highest 7.93% of the distribution for an individual player?
Solution :
Given that ,
mean = = 245
standard deviation = = 46
a.
P(x > 300) = 1 - P(x < 300)
= 1 - P[(x - ) / < (300 - 245) / 46)
= 1 - P(z < 1.20)
= 1 - 0.8849
= 0.1151
Probability = 0.1151
b.
P(x < 180) = P[(x - ) / < (180 - 245) / 46]
= P(z < -1.41)
= 0.0793
Probability = 0.0793
c.
P(190 < x < 320) = P[(190 - 245)/ 46) < (x - ) / < (320 - 245) / 46) ]
= P(-1.20 < z < 1.63)
= P(z < 1.63) - P(z <-1.20 )
= 0.9484 - 0.1151
= 0.8333
Probability = 0.8333
d.
= / n = 46 / 50 = 6.5054
P( > 260) = 1 - P( < 260)
= 1 - P[( - ) / < (260 - 245) / 6.5054]
= 1 - P(z < 2.31)
= 1 - 0.9896
= 0.0104
Probability = 0.0104
e.
Using standard normal table ,
P(Z < z) = 34.09%
P(Z < -0.41) = 0.3409
z = -0.41
Using z-score formula,
x = z * +
x = -0.41 * 46 + 245 = 226.14
weight = 226.14
f.
Using standard normal table ,
P(Z > z) = 7.93%
1 - P(Z < z) = 0.0793
P(Z < z) = 1 - 0.0793
P(Z < 1.41) = 0.9207
z = 1.41
Using z-score formula,
x = z * +
x = 1.41 * 46 + 245 = 309.86
weight = 309.86