Question 1205907: Flaws in plate glass used for large office buildings occur randomly at an average of 1 per 10 square feet.
PLEASE USE 4 DECIMAL PLACES IN THE ANSWERS.
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Part 1
a) What is the probability that a 6 ft. by 10 ft. sheet of this type of glass will contain less than 2 flaws?
The probability that a 6 ft. by 10 ft. sheet of this type of glass will contain less than 2 flaws is
enter your response here.
b) What is the probability that a 6 ft. by 10 ft. sheet of this type of glass will contain at least 1 flaw?
The probability that a 6 ft. by 10 ft. sheet of this type of glass will contain at least 1 flaw is
enter your response here.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! average number of flaws is 1 every 10 square feet.
if 60 square foot panel, then the average number of flaws is 6.
assuming a proportion type problem, then:
n = 60
p = .1
q = 1-p = .9
np = 60 * .1 = 6
s = sqrt(n*p*q) = sqrt(5.4) = 2.3238 rounded to 4 decimal places.
if average number of flaws is 2, then:
z = (2-6)/2.3238 = -1.72 rounded to 2 decimal places.
probability of getting a z-score less than that is .0427 rounded to 4 decimal places.
if average number of flaws is 1, then z = (1-6)/2.3238 = -2.15 rounded to 2 decimal places.
probability of getting a z-score greater than or equal to that is .9842 rounded to 4 decimal places.
these are ball park figures.
the actual value may not necessarily be exactly what they are, but they should be close.
best i can do based on what i know.
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