Question 596348
The weights of the contents of cans of tomato sauce produced by a company are normally distributed with a mean of 8 ounces and a standard deviation of 0.2 ounces.
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The measurement rod you must used is the z-distribution.
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a. What percentage of all cans produced contain more than 8.4 ounces of tomato paste?
z(8.4) = (8.4-8)/0.2 = 0.4/0.2 = 2
P(x > 8.4) = P(z > 2) = normalcdf(2,100) = 0.0228
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b. What percentage of all cans produced contain less than 7.8 ounces?
z(7.8) = (7.8-8)/0.2 = -1 
P(x < 7.8) = P(z < -1) = normalcdf(-100,-1) = 0.1587
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c. What percentage of cans contains between 7.4 and 8.2 ounces?
z(7.4) = (7.4-8)/0.2 = -3
z(8.2) = (8.2-8)/0.2 = 1
P(7.4< x <8.2) = P(-3< z <-1) = normcdf(-3,-1) = 0.1573
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d. Ninety-five percent of cans will contain at least how many ounces?
Find the z-value with a left-tail of 0.95:
invNorm(0.95) = 1.645
Find the corresponding "x" value:
x = zs+u
x = 1.645*0.2+8 = 8.3290
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e. What percentage of cans contains between 8.2 and 8.4 ounces?
Using my TI-84: normalcdf(8.2,8.4,8,0.2) = 0.1359
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Cheers,
Stan H.
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