Two independent vendors supply cement to a highway contractor. Through previous
ID: 3132086 • Letter: T
Question
Two independent vendors supply cement to a highway contractor. Through previous experience it is known that the compressive strength of samples of cement can be modeled by a normal distribution, with mu_1 = 6000 kilograms per square centimeter and sigma_1 = 100 kilograms per square centimeter for vendor 1, and mu_2 = 5825 and sigma_2 = 90 for vendor 2. What is the probability that both vendors supply a sample with compressive strength: Round your final answers to four decimal places (e.g. 98.7654). Less than 6100 kg/cm^2? Between 5800 and 6050? In excess of 6200? (Use Appendix A, Table I. Round value of standard normal random variable to 2 decimal places.)Explanation / Answer
a)
FOR VENDOR 1:
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 6100
u = mean = 6000
s = standard deviation = 100
Thus,
z = (x - u) / s = 1
Thus, using a table/technology, the left tailed area of this is
P(z < 1 ) = 0.8413
FOR VENDOR 2:
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 6100
u = mean = 5825
s = standard deviation = 90
Thus,
z = (x - u) / s = 3.06
Thus, using a table/technology, the left tailed area of this is
P(z < 3.06 ) = 0.9989
Hence,
P(both) = 0.8413*0.9989 = 0.8404 [ANSWER]
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b)
FOR VENDOR 1:
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 5800
x2 = upper bound = 6050
u = mean = 6000
s = standard deviation = 100
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -2
z2 = upper z score = (x2 - u) / s = 0.5
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.0228
P(z < z2) = 0.6915
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.6687
FOR VENDOR 2:
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 5800
x2 = upper bound = 6050
u = mean = 5825
s = standard deviation = 90
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.28
z2 = upper z score = (x2 - u) / s = 2.5
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.3897
P(z < z2) = 0.9938
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.6041
Hence,
P(both) = 0.6687*0.6041 = 0.4040 [ANSWER]
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c)
FOR VENDOR 1:
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 6200
u = mean = 6000
s = standard deviation = 100
Thus,
z = (x - u) / s = 2
Thus, using a table/technology, the right tailed area of this is
P(z > 2 ) = 0.0228
FOR VENDOR 2:
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 6200
u = mean = 5825
s = standard deviation = 90
Thus,
z = (x - u) / s = 4.17
Thus, using a table/technology, the right tailed area of this is
P(z > 4.17 ) = 0
Hence,
P(both) = 0.0228*0 = 0 [ANSWER]
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