Question

"Durable press" cotton fabrics are treated to improve their recovery from wrinkles after washing. Unfortunately, the treatment also reduces the strength of the fabric. The breaking strength of untreated fabric is normally distributed with mean 52 pounds and standard deviation 2.6 pounds. The same type of fabric after treatment has normally distributed breaking strength with mean 28.4 pounds and standard deviation 2 pounds. A clothing manufacturer tests 3 specimens of each fabric. All 6 strength measurements are independent. (Round your answers to four decimal places.)

a) What is the probability that the mean breaking strength of the 5 untreated specimens exceeds 50 pounds?

b) What is the probability that the mean breaking strength of the 5 untreated specimens is at least 25 pounds greater than the mean strength of the 5 treated specimens?

Answer 1

GIVEN:

The breaking strength of untreated fabric is normally distributed with mean pounds and standard deviation pounds.

The same type of fabric after treatment has normally distributed breaking strength with mean pounds and standard deviation pounds.

SOLUTION:

(a) PROBABILITY THAT THE MEAN BREAKING STRENGTH OF 5 UNTREATED SPECIMENS EXCEEDS POUNDS:

Since the specimens are independent, the sample mean of the breaking strengths of the 5 untreated specimens is normally distributed with mean pounds and standard deviation pounds.

So the probability that the sample mean of the breaking strengths of the 5 untreated specimens exceeds 50 pounds is

  

  

   {Since }

{From the z table, the probability value is the value corresponding to 1.7 row and 0.02 column.}

Thus the probability that the sample mean of the breaking strengths of the 5 untreated specimens exceeds 50 pounds is .

(b) PROBABILITY THAT THE MEAN BREAKING STRENGTH OF 5 UNTREATED SPECIMENS IS ATLEAST POUNDS GREATER THAN THE MEAN STRENGTH OF 5 TREATED SPECIMENS:

Since the specimens are independent, the sample mean of the breaking strength of the 5 untreated specimens is normally distributed with mean pounds and standard deviation pounds; the sample mean of the breaking strengths of the 5 treated specimens is normally distributed with mean pounds and standard deviation pounds.

Because the untreated and treated sample means are independent as well, the difference of the untreated minus treated sample means is normally distributed with mean pounds and standard deviation pounds.

(Recall that the variance of a sum or difference of *independent* random variables is the *sum* of their variances, so it follows that the standard deviation is the square root of the sum of the squares of their standard deviations.)

So the probability that the sample mean of the breaking strengths of the 5 untreated specimens is at least 25 pounds greater than the sample mean of the breaking strengths of the 5 treated specimens is

P(difference of the untreated minus treated sample means >= 25)

{Since }

{From the z table, the probability value is the value corresponding to 0.9 row and 0.05 column.}

Thus the probability that the sample mean of the breaking strengths of the 5 untreated specimens is at least 25 pounds greater than the sample mean of the breaking strengths of the 5 treated specimens is .