Question

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 π΅ and πΊ measure the weights of randomly selected 18 year old Statanadian boys and girls (π΅ for boys, πΊ for girls). What are the mean and standard deviation of the random variables π΅ and πΊ? b) Find the mean and standard deviation of the random variable π΅ β πΊ. Note that πΈ(π΅ β πΊ) = πΈ(π΅) β πΈ(πΊ) and πππ(π΅ β πΊ) = πππ(π΅) + πππ(πΊ). 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 π΅ β πΊ 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(π΅ β πΊ > 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?

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a) The mean of a random variable is denoted by π and the standard deviation is denoted by π. For the random variable π΅ (boys), the mean is 160 pounds and the standard deviation is 20 pounds. Thus, π_{π΅} = 160 and π_{π΅} = 20.

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

b) To find the mean and standard deviation of the random variable π΅ - πΊ, we can use the properties πΈ(π΅ - πΊ) = πΈ(π΅) - πΈ(πΊ) and πππ(π΅ - πΊ) = πππ(π΅) + πππ(πΊ).

The mean of π΅ - πΊ is π_{π΅ - πΊ} = π_{π΅} - π_{πΊ} = 160 - 135 = 25 pounds.

The standard deviation of π΅ - πΊ is π_{π΅ - πΊ} = β(π^2_{π΅} + π^2_{πΊ}) = β(20^2 + 17^2) β 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 π΅ - πΊ assigns to this pair is positive.

d) Since π΅ and πΊ are normally distributed and independent, π΅ - πΊ is also normally distributed with mean π_{π΅ - πΊ} and standard deviation π_{π΅ - πΊ}. We want to find P(π΅ - πΊ > 0).

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

π = (π΅ - πΊ - π_{π΅ - πΊ}) / π_{π΅ - πΊ}

Plugging in the values, we have:

π = (0 - 25) / 26.64 β -0.94

Now, we can find the probability using the standard normal distribution table.

P(π΅ - πΊ > 0) = P(π > -0.94)

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

Therefore, P(π΅ - πΊ > 0) β 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(π΅ - πΊ > 0), which we calculated to be approximately 0.8271.

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

b) To find the mean and standard deviation of the random variable π΅ - πΊ, we can use the properties πΈ(π΅ - πΊ) = πΈ(π΅) - πΈ(πΊ) and πππ(π΅ - πΊ) = πππ(π΅) + πππ(πΊ).

The mean of π΅ - πΊ is π_{π΅ - πΊ} = π_{π΅} - π_{πΊ} = 160 - 135 = 25 pounds.

The standard deviation of π΅ - πΊ is π_{π΅ - πΊ} = β(π^2_{π΅} + π^2_{πΊ}) = β(20^2 + 17^2) β 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 π΅ - πΊ assigns to this pair is positive.

d) Since π΅ and πΊ are normally distributed and independent, π΅ - πΊ is also normally distributed with mean π_{π΅ - πΊ} and standard deviation π_{π΅ - πΊ}. We want to find P(π΅ - πΊ > 0).

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

π = (π΅ - πΊ - π_{π΅ - πΊ}) / π_{π΅ - πΊ}

Plugging in the values, we have:

π = (0 - 25) / 26.64 β -0.94

Now, we can find the probability using the standard normal distribution table.

P(π΅ - πΊ > 0) = P(π > -0.94)

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

Therefore, P(π΅ - πΊ > 0) β 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(π΅ - πΊ > 0), which we calculated to be approximately 0.8271.

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