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 P(x)
0 0.0845
1 0.2700
2 0.3452
3 0.2207
4 0.0706
5 0.0090
To determine whether the table describes a probability distribution, we need to check if the probabilities sum up to 1 and if all the probabilities are non-negative.
Let's calculate the sum of the probabilities:
0.0845 + 0.2700 + 0.3452 + 0.2207 + 0.0706 + 0.0090 = 1
The sum of the probabilities is indeed equal to 1, so the table describes a probability distribution.
Now, let's find the mean and standard deviation:
To find the mean, multiply each x value by its corresponding probability and sum them up:
Mean = (0 * 0.0845) + (1 * 0.2700) + (2 * 0.3452) + (3 * 0.2207) + (4 * 0.0706) + (5 * 0.0090)
Mean = 0 + 0.2700 + 0.6904 + 0.6621 + 0.2824 + 0.0450
Mean = 1.9509
So, the mean of the probability distribution is 1.9509.
To find the standard deviation, we need to calculate the variance first. The formula for variance is:
Variance = Σ[(x - mean)^2 * P(x)]
where Σ denotes the sum.
First, calculate the squared deviation from the mean for each x value:
(0 - 1.9509)^2 * 0.0845 = 3.20214025 * 0.0845 = 0.2706197125
(1 - 1.9509)^2 * 0.2700 = 0.90369201 * 0.2700 = 0.244615467
(2 - 1.9509)^2 * 0.3452 = 0.02155304 * 0.3452 = 0.00744295568
(3 - 1.9509)^2 * 0.2207 = 2.37950521 * 0.2207 = 0.5253117627
(4 - 1.9509)^2 * 0.0706 = 4.83416001 * 0.0706 = 0.341294456
(5 - 1.9509)^2 * 0.0090 = 13.866169 * 0.0090 = 0.125795526
Now, sum up these squared deviations:
0.2706197125 + 0.244615467 + 0.00744295568 + 0.5253117627 + 0.341294456 + 0.125795526 = 1.51507988
Finally, calculate the square root of the variance to find the standard deviation:
Standard Deviation = √(Variance)
Standard Deviation = √(1.51507988)
Standard Deviation ≈ 1.2308
So, the mean of the probability distribution is 1.9509 and the standard deviation is approximately 1.2308.
To determine whether the table describes a probability distribution, we need to check if the probabilities add up to 1 and if all probabilities are non-negative.
Let's calculate the sum of the probabilities:
0.0845 + 0.2700 + 0.3452 + 0.2207 + 0.0706 + 0.0090 = 1
The sum of the probabilities is indeed equal to 1, which means that the table describes a probability distribution.
Now let's find the mean and standard deviation.
The mean (μ) can be calculated using the formula:
μ = ∑(x * P(x))
μ = (0 * 0.0845) + (1 * 0.2700) + (2 * 0.3452) + (3 * 0.2207) + (4 * 0.0706) + (5 * 0.0090)
Calculating the above expression:
μ = 0.2700 + 0.6904 + 0.6621 + 0.2828 + 0.2824 + 0.045 = 2.2327
So, the mean (expected value) of the random variable x is 2.2327.
To calculate the standard deviation (σ), we can use the formula:
σ = √[∑(x² * P(x)) - μ²]
σ = √[(0² * 0.0845) + (1² * 0.2700) + (2² * 0.3452) + (3² * 0.2207) + (4² * 0.0706) + (5² * 0.0090) - (2.2327)²]
Calculating the above expression:
σ = √[0 + 0.2700 + 0.6904 + 1.7379 + 0.2824 + 0.225 - 4.9853]
σ = √(2.9244)
σ ≈ 1.71
So, the standard deviation of the random variable x is approximately 1.71.
Therefore, the table describes a probability distribution with a mean of 2.2327 and a standard deviation of approximately 1.71.