Using the Z-score formula , 2-lb is the most that a bag of weigh and not need to repack.
Given that , the problem is normally distributed sample . For this we can use
Z-Score formula .
Z-Score formula for given mean value (μ) and standard deviations (σ ) is
Z =( X- μ )/ σ
Here , X is sample mean that is the bag which weigh and not need to be repacked .
From the given question, we have mean (μ) = 2-lb
Standard deviations (σ)= 0.05-lb
the Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score.
If bags are in the lower 4% implies 96% upper = 0.96 then p-value is 0.96..
By using Z-table we will get the value of Z-score corresponding to p-value .
Z-score value = 1.76
Put all the value in formula we get ,
1.76 = (X- 2)/0.05 => X-2 = 0.0880=> X = 2.088
Complete question :
An automatic packaging machine packages taffy into 2-lb bags. The weights of the bags are normally distributed with a mean of 2-lb and a standard deviation of 0. 05-lb. If bags are in the lower 4%, they are too light and must be repackaged. What is the most that a bag can weigh and not need to be repacked .
To learn more about Z-score formula , refer:
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