Question

When conducting research on color blindness in​ males, a researcher forms random groups with five males in each group. The random variable x is the number of males in the group who have a form of color blindness. 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 0.662 1 0.286 2 0.048 3 0.003 4 0.001 5 0.000

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

In order to check if we have a probability distirbution we need to satisfy some conditions:

1) $\sum_{i=1}^n P(X_i) = 1$

2) $P(X_i) \geq 0$

So then we satisfy the two conditions so then we have a probability distribution

$E(X) =\sum_{i=1}^n X_i P(X_i)$

$E(X)= 0*0.662 +1*0.286 +2*0.048 +3*0.003 + 4*0.001 +5*0 =0.395$

$E(X^2) =\sum_{i=1}^n X^2_i P(X_i)$

$E(X^2)= 0^2*0.662 +1^2*0.286 +2^2*0.048 +3^2*0.003 + 4^2*0.001 +5^2*0 =0.521$

$Var(X)= E(X^2) -[E(X)]^2$

$Var(X) = 0.521 -(0.395)^2 =0.365$

$Sd(X) = \sqrt{0.365}= 0.604$

Step-by-step explanation:

For this case we assume the following probability distribution:

X          0           1            2           3          4          5

P(X)  0.662   0.286   0.048   0.003   0.001   0.000

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.  

Solution to the problem

In order to check if we have a probability distirbution we need to satisfy some conditions:

1) $\sum_{i=1}^n P(X_i) = 1$

2) $P(X_i) \geq 0$

So then we satisfy the two conditions so then we have a probability distribution

For this case we can find the expected value with this formula:

$E(X) =\sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(X)= 0*0.662 +1*0.286 +2*0.048 +3*0.003 + 4*0.001 +5*0 =0.395$

In order to calculate the variance we need to calculate the second moment with this formula:

$E(X^2) =\sum_{i=1}^n X^2_i P(X_i)$

And replacing we got:

$E(X^2)= 0^2*0.662 +1^2*0.286 +2^2*0.048 +3^2*0.003 + 4^2*0.001 +5^2*0 =0.521$

And the variance is given by:

$Var(X)= E(X^2) -[E(X)]^2$

And replacing we got:

$Var(X) = 0.521 -(0.395)^2 =0.365$

And the standard deviation would be:

$Sd(X) = \sqrt{0.365}= 0.604$