let the random variable x denote the number of girls in a five-child family. if the probability of a female birth is .5
find the probability of 0,1,2,3,4, and 5 girls in a five-child family.
construct the binomial distribution and draw the histogram associated with the experiment
compute the mean and the standard deviation of the random variable x
for binomial coefficients use
C(n,k) = n!/(k![n-k]!)
here n = 5 and k is zero through five
or just use Pascal triangle
for n = 5
they are:
1 5 10 10 5 1
then
P(k girls) = C(n,k) p^k p^(5-k)
so for example
p(4 girls )= 5 (.5)^4 (.5)^1
= 5 (.5)^5 = .15625
The mean of a binomial distribution = n p so here = 5(.5) = 2.5
The sigma^2 = n p (1-p)
so here sigma^2 = 5(.5)(.5) = 5/4
so sigma = (1/2) sqrt 5 = 1.12
To find the probability of each possible outcome, we can use the binomial probability formula:
P(x) = (nCx) * p^x * (1-p)^(n-x)
Where:
- n is the number of trials (in this case, the number of children)
- x is the number of successes (in this case, the number of girls)
- p is the probability of success (in this case, the probability of a female birth)
- nCx is the combination formula for choosing x items out of n
Let's calculate the probabilities for each possible outcome:
For x = 0 (0 girls):
P(0) = (5C0) * (0.5^0) * (0.5^(5-0))
P(0) = 1 * 1 * 1/32
P(0) = 1/32
For x = 1 (1 girl):
P(1) = (5C1) * (0.5^1) * (0.5^(5-1))
P(1) = 5 * 0.5 * 0.5^4
P(1) = 5/32
For x = 2 (2 girls):
P(2) = (5C2) * (0.5^2) * (0.5^(5-2))
P(2) = 10 * 0.5^2 * 0.5^3
P(2) = 10/32
For x = 3 (3 girls):
P(3) = (5C3) * (0.5^3) * (0.5^(5-3))
P(3) = 10 * 0.5^3 * 0.5^2
P(3) = 10/32
For x = 4 (4 girls):
P(4) = (5C4) * (0.5^4) * (0.5^(5-4))
P(4) = 5 * 0.5^4 * 0.5^1
P(4) = 5/32
For x = 5 (5 girls):
P(5) = (5C5) * (0.5^5) * (0.5^(5-5))
P(5) = 1 * 0.5^5 * 0.5^0
P(5) = 1/32
Now, let's calculate the mean of the random variable x:
Mean (μ) = n * p
μ = 5 * 0.5
μ = 2.5
The standard deviation (σ) can be calculated using the formula:
Standard Deviation (σ) = √(n * p * (1 - p))
σ = √(5 * 0.5 * (1 - 0.5))
σ = √(5 * 0.5 * 0.5)
σ = √(1.25)
σ ≈ 1.12
To construct the binomial distribution and draw the associated histogram, we can plot the probabilities we calculated for each outcome on the y-axis and the number of girls (x) on the x-axis. We can use a bar chart to represent the probabilities visually.
To find the probability of each number of girls in a five-child family, we can use the binomial distribution formula. The binomial distribution is given by the formula:
P(x) = (nCx) * p^x * q^(n-x)
Where:
P(x) is the probability of getting exactly x girls
n is the total number of trials (in this case, the number of children, which is 5)
x is the number of successful trials (in this case, the number of girls)
p is the probability of success (the probability of a female birth, which is 0.5)
q is the probability of failure (the probability of a male birth, which is also 0.5)
nCx is the combination formula: n! / (x! * (n-x)!)
Now, let's calculate the probabilities for each number of girls in a five-child family:
For x = 0 (0 girls):
P(0) = (5C0) * (0.5)^0 * (0.5)^(5-0) = 1 * 1 * 0.03125 = 0.03125
For x = 1 (1 girl):
P(1) = (5C1) * (0.5)^1 * (0.5)^(5-1) = 5 * 0.5 * 0.03125 = 0.15625
For x = 2 (2 girls):
P(2) = (5C2) * (0.5)^2 * (0.5)^(5-2) = 10 * 0.25 * 0.03125 = 0.15625
For x = 3 (3 girls):
P(3) = (5C3) * (0.5)^3 * (0.5)^(5-3) = 10 * 0.125 * 0.03125 = 0.078125
For x = 4 (4 girls):
P(4) = (5C4) * (0.5)^4 * (0.5)^(5-4) = 5 * 0.0625 * 0.03125 = 0.015625
For x = 5 (5 girls):
P(5) = (5C5) * (0.5)^5 * (0.5)^(5-5) = 1 * 0.03125 * 1 = 0.03125
Now, let's construct the binomial distribution and draw the histogram:
The binomial distribution table for the experiment would look like this:
x P(x)
0 0.03125
1 0.15625
2 0.15625
3 0.078125
4 0.015625
5 0.03125
To draw the histogram, on the horizontal axis, we will label x as the number of girls, and on the vertical axis, we will label P(x) as the probability. We will represent the probabilities using bars of corresponding lengths for each x value.
The histogram would show a bar at x = 0 with a height of 0.03125, a bar at x = 1 with a height of 0.15625, a bar at x = 2 with a height of 0.15625, a bar at x = 3 with a height of 0.078125, a bar at x = 4 with a height of 0.015625, and a bar at x = 5 with a height of 0.03125.
Now, let's compute the mean and the standard deviation for the random variable x.
The mean (μ) of a binomial distribution is given by:
μ = n * p
In this case, μ = 5 * 0.5 = 2.5
The standard deviation (σ) of a binomial distribution is given by:
σ = sqrt(n * p * q)
In this case, σ = sqrt(5 * 0.5 * 0.5) = sqrt(1.25) = 1.118
Therefore, the mean of the random variable x is 2.5 and the standard deviation is 1.118.