Question

How to find the probability that the sample mean computed from a 25 measurements will exceed the sample mean computed from the 36 measurement by at least 3.4 but less than 5.9.?


A random sample of size 25 is taken from a normal population having a mean of 80 and a standard deviation of 5. A second sample of size 36 is taken from a different normal population having a mean of 75 and a standard deviation of 3. Find the probability that the sample mean computed from a 25 measurements will exceed the sample mean computed from the 36 measurement by at least 3.4 but less than 5.9. Assume the means to be measured to the nearest tenth.

Answer 1

`=> 58.34 %`

Explanation:

Values we are given:

`n_1 = 25, mu_1 = 80, sigma_1 = 5`
`n_2 = 36, mu_2 = 75, sigma_2 = 3`

First, we must find the probability of having a difference in means between the range `\Delta \barX_("lower")=3.4` and `\Delta \barX_("upper")=5.9`.

`P(\Delta \bar X_("lower") < \bar(X_1)-\bar(X_2) < \Delta \bar X_("upper"))`

`=>P(3.4 < \bar(X_1)-\bar(X_2) < 5.9)`

This probability is equivalent to the probability of finding a z-score `Z` within a range given by a lower bound of:

`(\Delta \bar X_("lower")-(mu_1-mu_2))/((sigma_1)/(sqrt(n_1)) + (sigma_2)/(sqrt(n_2)))`

`=>(3.4-(80-75))/((5)/(sqrt(25)) + (3)/(sqrt(36)))`

`=>color(blue)(-1.07)`

and an upper bound of:

`(\Delta \bar X_("upper")-(mu_1-mu_2))/((sigma_1)/(sqrt(n_1)) + (sigma_2)/(sqrt(n_2)))`

`=>(5.9-(80-75))/((5)/(sqrt(25)) + (3)/(sqrt(36)))`

`=>color(blue)(0.6)`

Hence, we can write:

`P(3.4 < \bar(X_1)-\bar(X_2) < 5.9) = P(color(blue)(-1.07) < Z < color(blue)(0.6))`

Now we have to look up probabilities from a z-score table. We need the negative z-score table for `-1.07` and the positive z-score table for `0.6`. We can find the values as shown below:

So now we have:

`P(3.4 < \bar(X_1)-\bar(X_2) < 5.9) = 0.7257-0.1423=color(green)(0.5834)`

Hence, the probability is `58.34 %`.