Let the random variable X denote the number of girls in a five child family. If the probability of a female birth is .5:
(a) Find the probabilities of 0, 1, 2, 3, 4, 5 girls in a five child family.
(b) Construct a binomial distribution and histogram associate with this experiment.
(c) Calculate the mean and standard deviation of the random variable X.
To find the probabilities of 0, 1, 2, 3, 4, and 5 girls in a five-child family, we can use the binomial probability formula.
(a) The binomial probability formula is given by:
P(x) = nCx * p^x * (1-p)^(n-x)
where:
- P(x) is the probability of getting exactly x girls,
- nCx is the number of ways to choose x objects out of n objects,
- p is the probability of a female birth, and
- (1-p) is the probability of a male birth.
For our case:
- n = 5 (number of children),
- x = 0, 1, 2, 3, 4, 5 (number of girls), and
- p = 0.5 (probability of a female birth).
Using the formula above, we can calculate the probabilities for each value of x:
P(0 girls) = 5C0 * 0.5^0 * (1-0.5)^(5-0)
P(1 girl) = 5C1 * 0.5^1 * (1-0.5)^(5-1)
P(2 girls) = 5C2 * 0.5^2 * (1-0.5)^(5-2)
P(3 girls) = 5C3 * 0.5^3 * (1-0.5)^(5-3)
P(4 girls) = 5C4 * 0.5^4 * (1-0.5)^(5-4)
P(5 girls) = 5C5 * 0.5^5 * (1-0.5)^(5-5)
Using the combinations formula (nCx = n! / (x! * (n-x)!)), we can calculate each probability:
P(0 girls) = 1 * 0.5^0 * (1-0.5)^(5-0)
P(1 girl) = 5 * 0.5^1 * (1-0.5)^(5-1)
P(2 girls) = 10 * 0.5^2 * (1-0.5)^(5-2)
P(3 girls) = 10 * 0.5^3 * (1-0.5)^(5-3)
P(4 girls) = 5 * 0.5^4 * (1-0.5)^(5-4)
P(5 girls) = 1 * 0.5^5 * (1-0.5)^(5-5)
Simplifying each calculation gives us the probabilities:
P(0 girls) = 0.03125
P(1 girl) = 0.15625
P(2 girls) = 0.3125
P(3 girls) = 0.3125
P(4 girls) = 0.15625
P(5 girls) = 0.03125
So, the probabilities of getting 0, 1, 2, 3, 4, and 5 girls in a five-child family are 0.03125, 0.15625, 0.3125, 0.3125, 0.15625, and 0.03125 respectively.
(b) To construct a binomial distribution and histogram for this experiment, we plot the number of girls (x) on the horizontal axis and the corresponding probabilities (P(x)) on the vertical axis. This results in a discrete probability distribution with bars representing each possible outcome.
The binomial distribution for this experiment would look like:
Number of Girls (x) Probability (P(x))
------------------------------------------
0 0.03125
1 0.15625
2 0.3125
3 0.3125
4 0.15625
5 0.03125
To create a histogram, we can represent each outcome (x) on the horizontal axis and the corresponding probability (P(x)) on the vertical axis, using bars to represent the probabilities. The height of each bar will be proportional to the probability of that outcome.
(c) To calculate the mean and standard deviation of the random variable X, we can use the formulas:
Mean (μ) = n * p
Standard Deviation (σ) = √(n * p * (1 - p))
For our case:
- n = 5 (number of children),
- p = 0.5 (probability of a female birth).
Calculating the mean:
Mean (μ) = 5 * 0.5 = 2.5
Calculating the standard deviation:
Standard Deviation (σ) = √(5 * 0.5 * (1 - 0.5))
= √(2.5 * 0.5)
= √(1.25)
= 1.118
So, the mean of the random variable X is 2.5 and the standard deviation is approximately 1.118.