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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], and the point estimate of the proportion of defective units is [tex]$\square$[/tex].



Answer :

GinnyAnswer
To solve this problem, we need to calculate two things: the point estimate of the population mean lifespan and the point estimate of the proportion of defective units.

1. Point Estimate of the Population Mean:

We calculate this by finding the average lifespan of the 40 sampled units. In this case, the calculated point estimate of the population mean is 32.3 days.

2. Point Estimate of the Proportion of Defective Units:

A unit is considered defective if its lifespan is less than 26 days. We must count how many units in the sample fall under this criterion and then find the proportion of these defective units in the total sample size of 40.

The number of defective units is 2, and hence the proportion of defective units in the sample is 2 / 40 = 0.05.

So, the completed answer is:
The point estimate of the population mean is [tex]$\boxed{32.3}$[/tex], and the point estimate of the proportion of defective units is [tex]$\boxed{0.05}$[/tex].

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