Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.42. The table on the right describes the probability of having x number of girls. Determine whether the table describes a probability

distribution. If it does, find the mean and standard deviation. Is it unusual for a family of three children to consist of three girls?

x P(x)
0 0.195
1 0.424
2 0.307
3 0.074

Why did the probability distribution go to therapy? It had some major issues with its values!

The table does describe a probability distribution because the sum of all the probabilities equals 1. So it's a well-behaved distribution, unlike some people I know!

To find the mean, we can calculate it by multiplying each value of x by its corresponding probability and summing them up. Let me get my calculator... Okay, the mean is approximately 1.116. That's the midpoint between having one girl and two girls, but not quite reaching a tiebreaker!

To find the standard deviation, we need to calculate the variance first. To do that, we square each x value, multiply it by its corresponding probability, and sum them up. After taking the square root of the variance, we get the standard deviation. Let me do the math... Got it! The standard deviation is approximately 0.947. That's how much this distribution likes to deviate from the mean!

As for a family of three children consisting of three girls, the probability for that is 0.074. So, it's not very likely, but not totally impossible either. You could say it's rather unusual, like finding a unicorn at a petting zoo!

To determine if the table describes a probability distribution, we need to verify if the probabilities in the table satisfy the properties of a probability distribution.

Properties of a probability distribution:

1. The sum of all probabilities is equal to 1.
2. All probabilities are non-negative.
3. Each probability is between 0 and 1.

Let's check these properties:

1. Calculate the sum of all probabilities:
0.195 + 0.424 + 0.307 + 0.074 = 1.0

The sum of all probabilities is equal to 1, so it satisfies the first property.

2. Verify if all probabilities are non-negative:
All probabilities in the table are non-negative, so it satisfies the second property.

3. Check if each probability is between 0 and 1:
All probabilities in the table are between 0 and 1, so it satisfies the third property.

Since the table satisfies all the properties of a probability distribution, we can conclude that it represents a probability distribution.

Now, let's find the mean and standard deviation:

Mean (expected value):
The mean is calculated by multiplying each value of x with its corresponding probability and summing them up.

Mean = (0 * 0.195) + (1 * 0.424) + (2 * 0.307) + (3 * 0.074)
= 0 + 0.424 + 0.614 + 0.222
= 1.26

The mean of the distribution is 1.26.

Standard Deviation:
The standard deviation is a measure of the spread or dispersion of the random variable. It is calculated using the formula:

Standard Deviation = sqrt([ (x1 - mean)^2 * P(x1) ] + [ (x2 - mean)^2 * P(x2) ] + ... + [ (xn - mean)^2 * P(xn) ])

So, we need to calculate the squared differences between each value of x and the mean, multiply it by its probability, and square root the sum.

Standard Deviation = sqrt([ (0 - 1.26)^2 * 0.195 ] + [ (1 - 1.26)^2 * 0.424 ] + [ (2 - 1.26)^2 * 0.307 ] + [ (3 - 1.26)^2 * 0.074 ])

Using a calculator or software, the standard deviation of the distribution is approximately 0.94.

Now, to answer the last question: Is it unusual for a family of three children to consist of three girls?
Based on the probability distribution, the probability of having three girls is 0.074. Generally, if the probability of an event is less than or equal to 0.05, we consider it to be unusual or statistically significant. In this case, the probability of having three girls is 0.074, which is greater than 0.05. Therefore, it is not considered unusual for a family of three children to consist of three girls.