SOLUTION: the weights of male basketball players on a certain camera are normally distributed with a mean of 180 punds and a standard deviation of 26 pounds. If a player is selected at rando
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Question 878221: the weights of male basketball players on a certain camera are normally distributed with a mean of 180 punds and a standard deviation of 26 pounds. If a player is selected at random find the probability that the player will weigh betwwn 180 and 225 pounds Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! Population: m = 180, sd = 26
P(180 < x < 225)= P(z = 45/26) - P(z = 0)= P(z = 1.7308) - P(z = 0)= .9583 - .5 = .4583 0r 45.83%
As You can see below, the Probability P(180 < x < 225) is expressed as the area
under the standard normal curve between z = 0 and z = .4583
To graph one would shade that area.
Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted along the Standard Normal Curve
Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50% to the right
