In the fictional country of Statanada, the weights of 18 year old boys are normally distributed with a mean of 160 pounds and a standard deviation of 20 pounds. The weights of 18 year old Statanadian girls are normally distributed with a mean of 135 pounds and a standard deviation of 17 pounds. a) Let the random variables 𝐵 and 𝐺 measure the weights of randomly selected 18 year old Statanadian boys and girls (𝐵 for boys, 𝐺 for girls). What are the mean and standard deviation of the random variables 𝐵 and 𝐺? b) Find the mean and standard deviation of the random variable 𝐵 − 𝐺. Note that 𝐸(𝐵 − 𝐺) = 𝐸(𝐵) − 𝐸(𝐺) and 𝑉𝑎𝑟(𝐵 − 𝐺) = 𝑉𝑎𝑟(𝐵) + 𝑉𝑎𝑟(𝐺). c) If a randomly selected 18 year old Statanadian boy weighs more than a randomly selected 18 year old Statanadian girl, what can you say about the value the random variable 𝐵 − 𝐺 assigns to this pair? d) The difference of two independent normally distributed random variables is another normally distributed random variable. Use this to find P(𝐵 − 𝐺 > 0). e) What is the probability that a randomly selected 18 year old Statanadian boy weighs more than a randomly selected 18 year old Statanadian girl?

a) The mean of a random variable is denoted by 𝜇 and the standard deviation is denoted by 𝜎. For the random variable 𝐵 (boys), the mean is 160 pounds and the standard deviation is 20 pounds. Thus, 𝜇_{𝐵} = 160 and 𝜎_{𝐵} = 20.

For the random variable 𝐺 (girls), the mean is 135 pounds and the standard deviation is 17 pounds. Thus, 𝜇_{𝐺} = 135 and 𝜎_{𝐺} = 17.

b) To find the mean and standard deviation of the random variable 𝐵 - 𝐺, we can use the properties 𝐸(𝐵 - 𝐺) = 𝐸(𝐵) - 𝐸(𝐺) and 𝑉𝑎𝑟(𝐵 - 𝐺) = 𝑉𝑎𝑟(𝐵) + 𝑉𝑎𝑟(𝐺).

The mean of 𝐵 - 𝐺 is 𝜇_{𝐵 - 𝐺} = 𝜇_{𝐵} - 𝜇_{𝐺} = 160 - 135 = 25 pounds.

The standard deviation of 𝐵 - 𝐺 is 𝜎_{𝐵 - 𝐺} = √(𝜎^2_{𝐵} + 𝜎^2_{𝐺}) = √(20^2 + 17^2) ≈ 26.64 pounds.

c) If a randomly selected 18 year old Statanadian boy weighs more than a randomly selected 18 year old Statanadian girl, the value the random variable 𝐵 - 𝐺 assigns to this pair is positive.

d) Since 𝐵 and 𝐺 are normally distributed and independent, 𝐵 - 𝐺 is also normally distributed with mean 𝜇_{𝐵 - 𝐺} and standard deviation 𝜎_{𝐵 - 𝐺}. We want to find P(𝐵 - 𝐺 > 0).

Using the properties of the standard normal distribution, we can standardize the random variable 𝐵 - 𝐺 by subtracting the mean and dividing by the standard deviation:

𝑍 = (𝐵 - 𝐺 - 𝜇_{𝐵 - 𝐺}) / 𝜎_{𝐵 - 𝐺}

Plugging in the values, we have:

𝑍 = (0 - 25) / 26.64 ≈ -0.94

Now, we can find the probability using the standard normal distribution table.
P(𝐵 - 𝐺 > 0) = P(𝑍 > -0.94)

Using the standard normal distribution table, we can find that P(𝑍 > -0.94) ≈ 0.8271.

Therefore, P(𝐵 - 𝐺 > 0) ≈ 0.8271.

e) The probability that a randomly selected 18 year old Statanadian boy weighs more than a randomly selected 18 year old Statanadian girl is equivalent to P(𝐵 - 𝐺 > 0), which we calculated to be approximately 0.8271.