A machine drills hole in a pipe with a mean diameter of 0.532 cm and a standard deviation of 0.002 cm. Calculate the control limits for mean of samples 5.
Question
A machine drills hole in a pipe with a mean diameter of 0.532 cm and a standard deviation of 0.002 cm. Calculate the control limits for mean of samples 5.
Solution
To calculate the control limits for the mean of samples of 5, we can use the formula:
Control limits = Mean ± (Z * (Standard Deviation / √Sample Size))
First, let's calculate the Z value. Since we want to calculate the control limits for a sample size of 5, we need to use the appropriate Z value for a sample size of 5.
The Z value can be found using a Z-table or a statistical software. For a sample size of 5, the Z value is typically 2.776.
Next, we can substitute the values into the formula:
Control limits = 0.532 cm ± (2.776 * (0.002 cm / √5))
Calculating the square root of 5, we get approximately 2.236.
Control limits = 0.532 cm ± (2.776 * (0.002 cm / 2.236))
Simplifying the equation:
Control limits = 0.532 cm ± (2.776 * 0.000894 cm)
Control limits = 0.532 cm ± 0.00248 cm
Therefore, the control limits for the mean of samples of 5 are approximately 0.52952 cm and 0.53448 cm.
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