Question

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. Determine whether a probability distribution is given. If a probability distribution is given, find its mean and standard deviation. If a probability distribution is not given, identify the requirements that are not satisfied.

x P(x)

0 0.028

1 0.153

2 0.319

3 0.319

4 0.153

5 0.028

Does the table show a probability distribution? Select all that apply.

A. Yes, the table shows a probability distribution.

B. No, the random variable x's number values are not associated with probabilities.

C. No, the random variable x is categorical instead of numerical.

D. No, not every probability is between 0 and 1 inclusive.

E. No, the sum of all the probabilities is not equal to 1.

Find the mean of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

A. muequals nothing child(ren) (Round to one decimal place as needed.)

B. The table does not show a probability distribution.

Find the standard deviation of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete your choice.

A. sigmaequals nothing child(ren) (Round to one decimal place as needed.)

B. The table does not show a probability distribution.

Answer #1

x | P(x) | |

0 | 0.028 | |

1 | 0.153 | |

2 | 0.319 | |

3 | 0.319 | |

4 | 0.153 | |

5 | 0.028 | |

SUM | 1 |

Yes, the table shows a probability distribution

why because

∑P(x) = 1 (Sum of probabilities)

and all probabilities are 0 < P(x) < 1

x | P(x) | x⋅p(x) | x^2⋅p(x) | ||

0 | 0.028 | 0 | 0 | ||

1 | 0.153 | 0.153 | 0.153 | ||

2 | 0.319 | 0.638 | 1.276 | ||

3 | 0.319 | 0.957 | 2.871 | ||

4 | 0.153 | 0.612 | 2.448 | ||

5 | 0.028 | 0.14 | 0.7 | ||

SUM | 1 | 2.5 | 7.448 | ||

μ=∑x⋅p(x) | 2.5 | σ = √(∑x^2⋅p(x)−μ^2) | 1.094531863 |

Mean μ = 2.5

Standard deviation σ = 1.1

Five males with an X-linked genetic disorder have one child
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five who inherit the X-linked genetic disorder. Determine whether
a probability distribution is given. If a probability distribution
is given, find its mean and standard deviation. If a probability
distribution is not given, identify the requirements that are not
satisfied.
x
P(x)
0
0.0290.029
1
0.1490.149
2
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3
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4
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5
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Five males with a particular genetic disorder have one child
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five who inherit the genetic disorder. Determine whether the table
describes a probability distribution. If it does, find the mean
and standard deviation.
x
0
1
2
3
4
5
P(x)
0.01160.0116
0.08340.0834
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x
0
1
2
3
4
5
P(x)
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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. Determine whether
a probability distribution is given. If a probability distribution
is given, find its mean and standard deviation. If a probability
distribution is not given, identify the requirements that are not
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0.029
Statistics

Five males with a particular genetic disorder have one child
each. The random variable x is the number of children among the
five who inherit the genetic disorder. Determine whether the table
describes a probability distribution. If it? does, find the mean
and standard deviation.
X 0 1 2 3 4 5
?P(x) 0.0053 0.0488 0.1811 0.3364 0.3124 0.1160

1. Use the given values of n and p to find the minimum usual
value
muμminus−2sigmaσ
and the maximum usual value
muμplus+2sigmaσ.
Round to the nearest hundredth unless otherwise noted.
nequals=10141014;
pequals=0.860.86
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. Determine whether
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