Question

A random sample of size n1=49 is selected from a distribution with a mean of 75 and a standard deviation of 7. A second random sample of size n2=36 is taken from another distribution with mean 70 and standard deviation 12. Let Xˉ1and Xˉ2 be the two sample means. Find:

(a) The probability distribution of X- 1 − Xˉ2 Explain.
(b) The probability that 3.5≤ Xˉ1 − Xˉ2 ≤5.5.
(c) Suppose that n 1 and n2 are unknown but it is given that n 1 +n 2 =150. Find the possible values of n 1 and n 2 if P{ Xˉ1 − Xˉ2 <6.3554}=0.7967.

Answer 1

a)
`\(\bar{X}_1 - \bar{X}_2\)` follows a normal distribution with mean 75 - 70 = 5 and standard deviation
`\(\sqrt{(7^2)/(49) + (12^2)/(36)}\).`

b) This probability is approximately 0.136, or 13.6%.

c) The values are
`n_1` = 60 and
`n_2` = 90. The sum is 150.

Step-by-step explanation:

(a) The probability distribution of
`\(\bar{X}_1 - \bar{X}_2\)` is normally distributed with a mean equal to the difference of the population means
`(\(\mu_1 - \mu_2\))`and a standard deviation equal to the square root of the sum of the variances divided by their respective sample sizes
`\((\sqrt{(\sigma_1^2)/(n_1) + (\sigma_2^2)/(n_2)})\). Thus, \(\bar{X}_1 - \bar{X}_2\)` follows a normal distribution with mean
`\(75 - 70 = 5\)` and standard deviation
`\(\sqrt{(7^2)/(49) + (12^2)/(36)}\).`

(b) To find the probability that
`\(3.5 \leq \bar{X}_1 - \bar{X}_2 \leq 5.5\)`, we standardize the values and then find the area under the standard normal curve between the z-scores corresponding to
`\(3.5\) and \(5.5\)`. Using the standard normal distribution table, we can calculate this probability.

(c) If
`\(n_1 + n_2 = 150\) and \(P\{ \bar{X}_1 - \bar{X}_2 < 6.3554\} = 0.7967\)`, we can determine the sample sizes
`\(n_1\) and \(n_2\)` by solving for them. By rearranging the formula for the standard deviation of
`\(\bar{X}_1 - \bar{X}_2\),` we can substitute in the given values and solve for
`\(n_1\) and \(n_2\).`