In a population, heights of males are normally distributed with u=180 cm and sigma^2=16 cm^2, while the heights of females are normally distributed with u=170 cm and sigma^2= 25 cm^2.

a) One random male and one random female are selected from the population
i)What is the probability that the male is more than 5 cm taller than the female?
ii)What is the distribution of the average of the two heights?

b)Now suppose that a random sample of 16 males and 16 females are selected from the
population.
i) What is the probability that the average height of the males is less than 178cm?
ii) What is the probability that the average height of the males is more than 12cm greater than the average height of the females?

So i said that X= height of one random male and Y= height of one random female. For this question I think I should find the z-score for both random variables and multiply the two probabilities since both variables are independent of each other. But I don't know how to get the probability of X+5>Y for a normal distribution and the other parts of the question, any help would be greatly appreciated, thanks.

To find the probability that a male is more than 5 cm taller than a female, you need to compare the two normal distributions for male and female heights. Since the heights are normally distributed, you can use the properties of the normal distribution to calculate the probability.

a) (i) Probability that the male is more than 5 cm taller than the female:
To solve this, you need to calculate the difference in means and variances for the male and female height distributions. Let's define the random variables X as the height of a random male and Y as the height of a random female.

Given:
For males: X ~ N(180, 16)
For females: Y ~ N(170, 25)

To find the probability that a male is more than 5 cm taller than a female, you can calculate the probability as P(X - Y > 5).

First, you need to convert the random variables X and Y into standard normal distribution variables (Z-scores) using the formula: Z = (X - μ) / σ, where Z ~ N(0,1).

So, for the male height random variable X:
μ_X = 180, σ_X = √16, and for a female height random variable Y:
μ_Y = 170, σ_Y = √25.

Calculate the Z-score for the male height:
Z_X = (X - μ_X) / σ_X = (X - 180) / √16

Similarly, calculate the Z-score for the female height:
Z_Y = (Y - μ_Y) / σ_Y = (Y - 170) / √25

Now, your original probability becomes: P(X - Y > 5) = P((X - Y) / √(16 + 25) > 5 / √(16 + 25)).

Since (X - Y) follows a normal distribution, the difference of two normal distributions is also normally distributed. The mean of the difference is μ_diff = μ_X - μ_Y = 180 - 170 = 10 and the variance is σ_diff^2 = σ_X^2 + σ_Y^2 = 16 + 25 = 41.

Therefore, you can calculate the probability using the standard normal distribution:

P((X - Y) / √(16 + 25) > 5 / √(16 + 25)) = P((X - Y) / √41 > 5 / √41).

Now, using standard normal tables or a calculator, look up the probability of Z > (5 / √41) and subtract it from 1 to find the final probability.

(ii) The distribution of the average of the two heights:
To find the distribution of the average of the male and female heights, let's define a new random variable A as the average height:

A = (X + Y) / 2

Since X and Y are normally distributed with means μ_X and μ_Y, respectively, and variances σ_X^2 and σ_Y^2, the sum X + Y is also normally distributed with a mean of μ_X + μ_Y and a variance of σ_X^2 + σ_Y^2.

So, the mean of the average height A is (μ_X + μ_Y) / 2 and the variance is (σ_X^2 + σ_Y^2) / 4.