Answer :
Answer:
c)
The probability of a player weighing more than 238
P( X > 238) = 0.0174
Step-by-step explanation:
Step(i):-
Given mean of the normally distribution = 195 pounds
Given standard deviation of the normally distribution
= 20 pounds.
Let 'x' be the random variable of the normally distribution
Let X = 238
[tex]Z = \frac{x-mean}{S.D} = \frac{238-195}{20} = 2.15[/tex]
Step(ii):-
The probability of a player weighing more than 238
P( X > 238) = P( Z> 2.15)
= 1 - P( Z < 2.15)
= 1 - ( 0.5 + A(2.15)
= 1 - 0.5 - A(2.15)
= 0.5 - 0.4821 ( from normal table)
= 0.0174
The probability of a player weighing more than 238
P( X > 238) = 0.0174
