
| x | f(x) | xP(x) | x2P(x) |
| 0 | 0.021 | 0.000 | 0.000 |
| 1 | 0.121 | 0.121 | 0.121 |
| 2 | 0.284 | 0.568 | 1.135 |
| 3 | 0.333 | 1.000 | 2.999 |
| 4 | 0.196 | 0.782 | 3.130 |
| 5 | 0.046 | 0.230 | 1.148 |
| total | 2.700 | 8.532 | |
| E(x) =μ= | ΣxP(x) = | 2.7000 | |
| E(x2) = | Σx2P(x) = | 8.5320 | |
| Var(x)=σ2 = | E(x2)-(E(x))2= | 1.2420 | |
| std deviation= | σ= √σ2 = | 1.1145 |
from above mean μ =2.700
and std deviation σ=1.1
4. Five males with a particular genetic disorder have onc child each. I need the second...
4.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.
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. μ =
(Round to one decimal place as needed.)...
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) 0.4182 0.3983 0.1517 0.0289 0.0028 0.0001 Find the mean of the random variable x. Select the correct choice below and, if necessary, fill in the answer box...
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ive 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 0.0290.029 1 0.1550.155 2 0.3160.316 3 0.3160.316 4 0.1550.155 5 0.0290.029 Does the table show...