Question 673195: the amount of chemical W used to manufacture certain detergents vary, depending on the concentration and use of the required detergent. company records show that the amounts used are normally distributed. futher, they indicate that the chances of any detergent having at least 2.4grams of this chemical are 2.28% while the chances of having less than 1.4grams are 30.85%
a) determine the probability of a detergent having
i)more than 1.2grams but not exceeding 2.6grams of the chemical
ii)either less than 0.4grams, exactly 0.8grams or more than 2.8grams
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! the chances of any detergent having at least 2.4grams of this chemical are 2.28% while the chances of having less than 1.4grams are 30.85%
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The z-value with a right tail of 2.28% is z = invNorm(0.9772) = 1.9991
The z-value with a left tail of 30.85% is z = invNorm(0.3085) = 0.5001
Equations based on x = z*s + u
2.4 = 1.9991*s + u
1.4 = 0.5001*s + u
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Subtract to solve for "s":
1 = 1.499s
s = 0.667
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Solve for "u":
2.4 = 1.9991*0.667 + u
u = 1.0666
a) determine the probability of a detergent having
i)more than 1.2grams but not exceeding 2.6grams of the chemical
z(1.2) = (1.2-1.0666)/0.667 = 0.1999
z(2.6) = (2.6-1.0666)/0.667 = 2.2990
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P(1.2 < x < 2.6) = P(0.1999 < z < 2.2990) = 0.4100
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ii)
either less than 0.4grams
z(0.4) = (0.4-1.0666)/0.667 = -0.999
P(x < 0.4) = P(z < -0.999) = normalcdf(-100,-0.999) = 0.1589
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exactly 0.8grams
P(x = 0.8) = 0
Why?:: The probability of ANY particular value in a continuous
distribution is zero.
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more than 2.8grams
z(2.8) = (2.8-1.0666)/0.667 = 2.5987
P(x > 2.8) = P(z > 2.5987) = normalcdf(2.5987,100) = 0.0047
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cheers,
Stan H.
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