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.

1 Answer
Mar 25, 2018

#=> 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:
ttable
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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 %#.