binomial distribution

Sue0113

Junior Member
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Feb 1, 2012
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Consider a basketball game where a player throws from the free throw line. Her success rate is 65%. If she throws 10 times in a game determine the following:The mean and standard deviation of this binomial distribution.
p=.65 and q =.35 and n=10
Mean is np
.65 X 10= 6.5
Therefore mean is 6.5
SD = the square root of (npq)
.65 X 10 X .35 =2.275
Therefore Standard deviation is 2.275
Is this correct?
 
Consider a basketball game where a player throws from the free throw line.
Her success rate is 65%. If she throws 10 times in a game determine the following:The mean and
standard deviation of this binomial distribution.

p=.65 and q =.35 and n=10
Mean is np
.65 X 10= 6.5
Therefore mean is 6.5 . . . Correct.


SD = the square root of (npq)
.65 X 10 X .35 =2.275
Therefore Standard deviation is 2.275 . . . Two lines above you mentioned that the square root must be taken.
Is this correct?

So, compute \(\displaystyle \sqrt{2.275 \ \ } \) to get the standard deviation.
 
Then the sd is

square root of 2.275 = 1.508
Therefore the Standard deviation is 1.508.
 
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