SOLUTION: Can someone help me with this question, please? Sugar packaged by a certain machine has a mean weight of 5 lb and a standard deviation of 0.03 lb. For what values of c can the m

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Question 927549: Can someone help me with this question, please?
Sugar packaged by a certain machine has a mean weight of 5 lb and a standard deviation of 0.03 lb. For what values of c can the manufacturer of the machinery claim that the sugar packaged by this machine has a weight between 5 − c and 5 + c lb with probability at least 91%?
at least c =
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
you are asked to find c such that, if W+denotes the weight of a package of sugar,
P%285-c+%3C+W+%3C5%2Bc+%29%3C=0.91
Since W is normally distributed, transform it into a *standard* normally distributed variable, Z, using the formula:
Z=(W−μ)σ <=> W=Zσ+μ
where μ and σ are the mean and std. deviation of W's distribution.
P%285-c+%3C0.03Z%2B5+%3C5%2Bc%29+%3C=0.91
P%28-c%2A0.03%3CZ%3Cc%2A0.03%29%3C=0.91
P%28-c%2A0.03%3CZ%3Cc%2A0.03%29%3C=0.91

Due to the symmetry of the normal curve, you can write this as
P%280%3CZ%3Cc%2A0.03%29%3C=0.455
P%28Z%3Cc%2A0.03%29-P%28Z%3C0%29%3C=0.455
P%28Z%3Cc%2A0.03%29-0.5%3C=0.455
P%28Z%3Cc%2A0.03%29%3C=0.955
Referring to a z-table, you get a corresponding z-value of 1.70
In other words,
P%28Z%3Cc%2A0.03%29%3C=0.955 <=> P%28Z%3C1.70%29%3C=0.955
which means 1.7=c%2A0.03 => c=0.051