Question

Asked Nov 27, 2019

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Step 1

Hey, since there are multiple subpart questions posted, we will answer first three subpart questions. If you want any specific subpart question to be answered then please submit that subpart question only or specify the subpart question number in your message.”

Step 2

**Solution:**

**a.**

**The number of shirts is obtained below:**

The population mean is 15.5 inches and standard deviation is 0.5 inches. The total number of shirts in a batch is 1,500.

From the Excel,

**“=NORM.DIST(14,15.5,0.5,TRUE)]” **in EXCEL software.

The probability -value is **0.0013.**

The required value=1,500x0.0013= 2.0248.

The number of shirts is** 2.0248.**

Step 3

**b.**

**The number of shirts is obtained below:**

From the Excel,

**“=NORM.DIST(14.5,15.5,0.5,TRUE)]” **in EXCEL software.

The probability -value is **0.0228.**

The required value=1,500x0.0228= 34.125.

The number of shirts is** 34.125.**

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