In: Statistics and Probability
"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 1.8 pounds. The same type of fabric after treatment has normally distributed breaking strength with mean 24.1 pounds and standard deviation 1.8 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 3 untreated specimens exceeds 50 pounds? (b) What is the probability that the mean breaking strength of the 3 untreated specimens is at least 25 pounds greater than the mean strength of the 3 treated specimens?
a) Normal Distribution with mean 52 and standard deviation 1.8 looks like the distribution shown where probability of a fabric having strength greater than 50 pounds is the shaded area shown under the curve.
So, for independent untreated fabric specimen selections the probability of having three specimens with mean strength greater than 50 pounds will be same as probability of selecting a specimen with strength greater than 50 pounds which can be calculated from Z-score for the distribution -
z = (50-52)/1.8
z = 1.1111
From z-score table we can get
probability = 0.8667 that is 86.67%
b) For calculation the probability for the difference between means of selections from treated and untreated specimen we can have a look at the distribution of difference between two normal distributions
which would have a mean of μ=μ2−μ1 and variance σ2=σ12+σ22
mean = 52- 24.1 = 27.9
standard deviation = (1.82 + 1.82)1/2 = 1.8
Probability of having a difference in means greater than 25 is the shaded area
z-score for the area is (27.9-25)/1.8 = 1.6111
probability of having a difference of at least 25 pounds in mean strength (from z-table) is 0.9464
that is 94.64%.