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A company manufactures 2,000 units of its flagship product in a day. The quality control department takes a random sample of 40 units to test for quality. The product is put through a wear-and-tear test to determine the number of days it can last. If the product has a lifespan of less than 26 days, it is considered defective. The table gives the sample data that a quality control manager collected.

\begin{tabular}{|l|l|l|l|l|}
\hline 39 & 31 & 38 & 40 & 29 \\
\hline 32 & 33 & 39 & 35 & 32 \\
\hline 32 & 27 & 30 & 31 & 27 \\
\hline 30 & 29 & 34 & 36 & 25 \\
\hline 30 & 32 & 38 & 35 & 40 \\
\hline 29 & 32 & 31 & 26 & 26 \\
\hline 32 & 26 & 30 & 40 & 32 \\
\hline 39 & 37 & 25 & 29 & 34 \\
\hline
\end{tabular}

The point estimate of the population mean is [tex]$\square$[/tex].
The point estimate of the proportion of defective units is [tex]$\square$[/tex].


Sagot :

To determine the point estimate of the population mean and the point estimate of the proportion of defective units, we need to process the data provided in the sample. Here is the step-by-step solution:

1. Find the Point Estimate of the Population Mean:

The point estimate of the population mean is the average lifespan of the units in the sample. Summing up all the lifespans and dividing by the number of units in the sample gives us the mean.

Given the result from the earlier analysis:
- The point estimate of the population mean is 32.3 days.

2. Find the Point Estimate of the Proportion of Defective Units:

A unit is considered defective if its lifespan is less than 26 days. We need to count the number of defective units and divide this by the total number of units in the sample to obtain the proportion of defective units.

Given the result from the earlier analysis:
- The point estimate of the proportion of defective units is 0.05 (or 5%).

Therefore, the point estimate of the population mean is 32.3, and the point estimate of the proportion of defective units is 0.05.