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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 𝐡 and 𝐺 measure the weights of randomly selected 18 year old Statanadian boys and girls (𝐡 for boys, 𝐺 for girls). What are the mean and standard deviation of the random variables 𝐡 and 𝐺? b) Find the mean and standard deviation of the random variable 𝐡 βˆ’ 𝐺. Note that 𝐸(𝐡 βˆ’ 𝐺) = 𝐸(𝐡) βˆ’ 𝐸(𝐺) and π‘‰π‘Žπ‘Ÿ(𝐡 βˆ’ 𝐺) = π‘‰π‘Žπ‘Ÿ(𝐡) + π‘‰π‘Žπ‘Ÿ(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 𝐡 βˆ’ 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(𝐡 βˆ’ 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?

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Answer to a math 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 𝐡 and 𝐺 measure the weights of randomly selected 18 year old Statanadian boys and girls (𝐡 for boys, 𝐺 for girls). What are the mean and standard deviation of the random variables 𝐡 and 𝐺? b) Find the mean and standard deviation of the random variable 𝐡 βˆ’ 𝐺. Note that 𝐸(𝐡 βˆ’ 𝐺) = 𝐸(𝐡) βˆ’ 𝐸(𝐺) and π‘‰π‘Žπ‘Ÿ(𝐡 βˆ’ 𝐺) = π‘‰π‘Žπ‘Ÿ(𝐡) + π‘‰π‘Žπ‘Ÿ(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 𝐡 βˆ’ 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(𝐡 βˆ’ 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?

Expert avatar
Brice
4.8
113 Answers
a) The mean of a random variable is denoted by πœ‡ and the standard deviation is denoted by 𝜎. For the random variable 𝐡 (boys), the mean is 160 pounds and the standard deviation is 20 pounds. Thus, πœ‡_{𝐡} = 160 and 𝜎_{𝐡} = 20.

For the random variable 𝐺 (girls), the mean is 135 pounds and the standard deviation is 17 pounds. Thus, πœ‡_{𝐺} = 135 and 𝜎_{𝐺} = 17.

b) To find the mean and standard deviation of the random variable 𝐡 - 𝐺, we can use the properties 𝐸(𝐡 - 𝐺) = 𝐸(𝐡) - 𝐸(𝐺) and π‘‰π‘Žπ‘Ÿ(𝐡 - 𝐺) = π‘‰π‘Žπ‘Ÿ(𝐡) + π‘‰π‘Žπ‘Ÿ(𝐺).

The mean of 𝐡 - 𝐺 is πœ‡_{𝐡 - 𝐺} = πœ‡_{𝐡} - πœ‡_{𝐺} = 160 - 135 = 25 pounds.

The standard deviation of 𝐡 - 𝐺 is 𝜎_{𝐡 - 𝐺} = √(𝜎^2_{𝐡} + 𝜎^2_{𝐺}) = √(20^2 + 17^2) β‰ˆ 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 𝐡 - 𝐺 assigns to this pair is positive.

d) Since 𝐡 and 𝐺 are normally distributed and independent, 𝐡 - 𝐺 is also normally distributed with mean πœ‡_{𝐡 - 𝐺} and standard deviation 𝜎_{𝐡 - 𝐺}. We want to find P(𝐡 - 𝐺 > 0).

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

𝑍 = (𝐡 - 𝐺 - πœ‡_{𝐡 - 𝐺}) / 𝜎_{𝐡 - 𝐺}

Plugging in the values, we have:

𝑍 = (0 - 25) / 26.64 β‰ˆ -0.94

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

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

Therefore, P(𝐡 - 𝐺 > 0) β‰ˆ 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(𝐡 - 𝐺 > 0), which we calculated to be approximately 0.8271.

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