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 |
0.035 |
||||

1 |
0.157 |
||||

2 |
0.308 |
||||

3 |
0.308 |
||||

4 |
0.157 |
||||

5 |
0.035 |

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

A.

Yes, the table shows a probability distribution.

B.

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

C.

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

D.

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

E.

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

Answer #1

Solution :

The sum of the probability is equal to 1 .

P(X) = 1

A.

Yes, the table shows a probability distribution.

x | P(x) | x * P(x) |
x^{2} * P(x) |

0 | 0.035 | 0 | 0 |

1 | 0.157 | 0.157 | 0.157 |

2 | 0.308 | 0.616 | 1.232 |

3 | 0.308 | 0.924 | 2.772 |

4 | 0.157 | 0.628 | 2.512 |

5 | 0.035 | 0.175 | 0.875 |

Sum | 1 | 2.5 | 7.548 |

Mean = = X * P(X) = 2.5

Standard deviation =

=X
^{2} * P(X) -
^{2}

= 7.548
- 2.5^{2}

= 1.1393

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five who inherit the X-linked genetic disorder. Determine whether
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