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1 answer:
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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:
d = 234.6 m
Step-by-step explanation:
You can consider a system of coordinates with its origin at the beginning of the walk of the student.
When she start to walk, she is at (0,0)m. After her first walk, her coordinates are calculated by using the information about the incline and the distance that she traveled:
she is at the coordinates (52.97 , 28.16)m.
Next, when she walks 180m to the east, her coordinates are:
(52.97+180 , 28.16)m = (232.97 , 28.16)m
To calculate the distance from the final point of the student to the starting point you use the Pythagoras generalization for the distance between two points:
The displacement of the student on her complete trajectory was of 234.6m