Its 148 cm2
S = 2(l*w + l*h + w*h)
Answer and Explanation:
Given : Five males with an X-linked genetic disorder have one child each. The random variable x is the number of children among the five who inherit the X-linked genetic disorder.
To find :
a. Does the table show a probability distribution?
b. Find the mean and standard deviation of the random variable x.
Solution :
a) To determine that table shows a probability distribution we add up all six probabilities if the sum is 1 then it is a valid distribution.


Yes it is a probability distribution.
b) First we create the table as per requirements,
x P(x) xP(x) x² x²P(x)
0 0.029 0 0 0
1 0.147 0.147 1 0.147
2 0.324 0.648 4 1.296
3 0.324 0.972 9 2.916
4 0.147 0.588 16 2.352
5 0.029 0.145 25 0.725
∑P(x)=1 ∑xP(x)=2.5 ∑x²P(x)=7.436
The mean of the random variable is

The standard deviation of the random sample is







Therefore, The mean is 2.5 and the standard deviation is 1.08.
Answer: The mean increases by 3
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The original data set is
{50, 76, 78, 79, 79, 80, 81, 82, 82, 83}
The outlier is 50 because it is not near the group of values from 76 to 83 which is where the main cluster is.
The original mean is M = (50+76+78+79+79+80+81+82+82+83)/10 = 77
If we take out the outlier 50, the new mean is N = (76+78+79+79+80+81+82+82+83)/9 = 80
So in summary so far
old mean = M = 77
new mean = N = 80
The difference in values is N-M = 80-77 = 3
So that's why the mean increases by 3