A random sample of size n1 = 16 is selected from a normal population with a mean of 75 and standard deviation of 8. A second random sample of size n2 = 9 is taken independently from another normal population with mean 70 and standard deviation of 12. Let X1 and X2 be the two sample means. Find
(a) The probability that X1 − X2 exceeds 4.
(b) The probability that 3.5 < X1 − X2 < 5.5.

The sum of independent normally distributed random variable is also a normally distributed random variable. The needed probabilities are given as:

(a) The probability that X1 − X2 exceeds 4 is 0.5279 approx
(b) The probability that 3.5 < X1 − X2 < 5.5 is 0.0836 approx

What is the distribution of random variable which is sum of normal distributions?

Suppose that a random variable X is formed by n mutually independent and normally distributed random variables such that:

[tex]X_i = N(\mu_i , \sigma^2_i) ; \: i = 1,2, \cdots, n[/tex]

And if

[tex]X = X_1 + X_2 + \cdots + X_n[/tex]

Then, its distribution is given as:

[tex]X \sim N(\mu_1 + \mu_2 + \cdots + \mu_n, \: \: \sigma^2_1 + \sigma^2_2 + \cdots + \sigma^2_n)[/tex]

For the given case, let we take:

[tex]X_1[/tex] = Random variable assuming values of sample 1
[tex]X_2[/tex] = Random variable assuming values of sample 2

Then, we have:

[tex]X_1 \sim N(\mu = 75, \sigma^2 = (8)^2 = 64)\\\\X_2 \sim N(\mu = 70, \sigma^2 = (12)^2 = 144)\\[/tex]

Then, The random variable [tex]-X_2[/tex] has all negative values than of [tex]X_2[/tex], so its mean will also become negative, but standard deviation would be same(since its measure of spread which would be same for [tex]-X_2[/tex] .

Thus, [tex]-X_2 \sim N(-70, 12^2) = N(-70, 144)[/tex]

Thus, we get:

[tex]X = X_1 - X_2\\\\X \sim N(75 -70, \sigma^2 = 64 + 144)\\\\X \sim N(5, 208)[/tex]

Thus, mean of X is 5, and standard deviation is

[tex]\sqrt{208} \approx 14.42[/tex]

(positive root since standard deviation is non negative quantity)

Thus, calculating the needed probability with the use of z-scores:

Case 1: P( X1 − X2 exceeds 4)

P(X > 4)

Converting to standard normal distribution, we get

[tex]Z = \dfrac{X - \mu}{\sigma} \approx \dfrac{X - 5}{14.42}\\\\P(X > 4) \approx P(Z > \dfrac{4-5}{14.42}) \approx P(Z > -0.07) = 1-P(Z \leq -0.07)\\[/tex]

Using the z-tables, the p value for z = -0.07 is 0.4721

p value for Z = z gives [tex]P(Z \leq z) = p[/tex]

And therefore,

[tex]P(Z \leq -0.07) = 0.4721\\\\\rm and\: thus\\\\P(X > 4) = P(Z > -0.07) = 1 - P(Z \leq -0.07) = 1 - 0.4721 = 0.5279[/tex]

b) The probability that 3.5 < X1 − X2 < 5.5.

P( 3.5 < X1 − X2 < 5.5 ) = P( 3.5 < X < 5.5)

Using z scores, we get:

[tex]P( 3.5 < X < 5.5) = P(X < 5.5) - P(X < 3.5) \\\\P(3.5 < X < 5.5) \approx P(Z < \dfrac{5.5-5}{14.2}) - P(Z < \dfrac{3.5 - 5}{14.2})\\\\P(3.5 < X < 5.5) \approx P(Z < 0.113) - P(Z < -0.104)[/tex]

Using the z-tables, p value for z = 0.113 is 0.5438

and p value for z = -0.104 is 0.4602
Thus,

[tex]P(3.5 < X < 5.5) \approx P(Z < 0.113) - P(Z < -0.104)\\P(3.5 < X < 5.5) \approx 0.5438 - 0.4602 = 0.0836[/tex]

Learn more about standard normal distribution here:

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