Using the normal probability distribution principle given the mean and standard deviation of Buckley's farm ; the solutions to the exercise are :
- P(Z < z) = 10.6%
- Mean = 106.9 ounces ; Standard deviation = 0.298 ounces
- Mean weight that should be used = 15.95 ounces
Given the Parameters :
- Mean, μ = 16.15 ounces
- Standard deviation, σ = 0.12 ounces
Recall :
- Zscore = \(\frac{\bar{x} - \mu}{σ}\)
Zscore = \(\frac{16 - 16.15}{0.12} = -1.25\)
P(Z < - 1.25) = 0.1056 = 10.6% (normal distribution table)
2.)
Empty box :
- Weight = 10 ounces ; Standard deviation = 0.05 ounces
Mean weight of 6 bags :
10 + (6 × 16.15)
10 + 96.90 = 106.90 ounces
Standard deviation of 6 bags :
√(0.05² + 0.12² + 0.12² + 0.12² + 0.12² + 0.12² + 0.12²) = 0.298 ounces
3.)
Using the normal distribution table ;
Zscore for 5% = 0.05 = - 1.645
Using the Zscore formula :
\(-1.645 = \frac{x - 16.15}{0.12}\)
Cross multiply :
x - 16.15 = -1.645 × 0.12
x = - 19.74 + 16.15
x = 15.95
Therefore, the mean weight that should be used is 15.95 ounces
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