Paint Durability. A paint manufacturer in Pittsburgh claims that his paint will last an average of $5$ years. Assuming that paint life is normally distributed and has a standard deviation of $0.5$ year. answer the following questions:

a. Suppose that you paint one house with the paint and that the paint lasts $4.5$ years. Would you consider that evidence against the manufacturer's claim? (Hint: Assuming that the manufacturer's claim is correct, determine the probability that the paint life for a randomly selected house painted with the paint is $4.5$ years or less.)

b. Suppose that you paint $10$ houses with the paint and that the paint lasts an average of $4.5$ years for the $10$ houses. Would you consider that evidence against the manufacturer's claim?

c. Repeat part (b) if the paint lasts an average of $4.9$ years for the $10$ houses painted.

A Pittsburgh paint maker promises that his paint will last on average $5$ years. Paint life is regularly distributed with a $0.5$ year standard deviation.

The formula used:${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{10}}$

Let $x$be the paint life (in years) of a randomly selected house.

$x$is normally distributed, according to the manufacturer, with a mean of $\mu =5$years and a standard deviation of$\sigma =0.5$ years.

Assuming manufacturer's claim is correct,

$$\begin{gathered} P(x\le 4.5) \\ =P\left(\frac{x-\mu }{\sigma }\le \frac{4.5-\mu }{\sigma }\right) \\ =P\left(z\le \frac{4.5-5}{0.5}\right) \\ =P(z\le -1) \\ =0.1587 \end{gathered}$$

As a result, we cannot use that data to refute the manufacturer's claim in the current scenario. Because, assuming the manufacturer's promise is right, the paint life for a randomly picked house painted with this paint is $0.1587$ i.e., such an event would occur about $16\%$ of the time.

Let $\overline{x}$be the mean paint life for 10 randomly selected house. $\overline{x}=4.5$Years

$\therefore$According to manufacturer's claim, $\overline{x}~N\left(5,{\sigma }_{\overline{x}}^2\right)$

where ${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{10}}$

$$\begin{gathered} =\frac{0.5}{\sqrt{10}} \\ =0.158 \end{gathered}$$

$$\begin{gathered} P(\overline{x}\le 4.5) \\ =P\left(\frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{4.5-\mu }{{\sigma }_{\overline{x}}}\right) \\ =P\left(z\le \frac{4.5-5}{0.158}\right) \\ =P(z\le -3.165) \\ =0.0008 \end{gathered}$$

As a result, we can analyze the evidence against the manufacturer's claim in the current situation. Because, assuming the manufacturer's claim is right, the average paint life for ten randomly selected buildings coated with this paint is $0.0008$, implying that such an event occurs less than$0.1\%$ of the time.

The probability $P(z\le -3.165)$is calculated using MINITAB in the following way:

Step 1: Press the Calc menu; Highlight the 'Probability Distributions'.

Step 2: Press Normal... ;

Step 3: Tick $\square$Cumulative Probability and enter the given data values

Mean: $0$

Standard deviation: $1$

Step 4: Tick $\square$Input constant and enter the value $-3.165$

- Input constant: $-3.165$

Step 5: Press Ok

Given that $\overline{x}=4.9$Years

According to the manufacturer's claim, $\overline{x}~N\left(5,{\sigma }_{\overline{x}}^2\right)$, where ${\sigma }_{\overline{x}}=0.158$

$$\begin{gathered} P(\overline{x}\le 4.9) \\ =P\left(\frac{\overline{x}-\mu }{{\sigma }_{\overline{x}}}\le \frac{4.9-\mu }{{\sigma }_{\overline{x}}}\right) \\ =P\left(z\le \frac{4.9-5}{0.158}\right) \\ =P(z\le -0.633) \\ =0.263367 \end{gathered}$$

As a result, we cannot use that data to refute the manufacturer's claim in the current scenario. Because, assuming the manufacturer's claim is right, the average paint life for ten randomly selected houses painted with this paint is $0.263367$, implying that such an event occurs roughly $26\%$ of the time.

The probability, $P(z\le -0.633)$ is calculated using MINITAB in the follwing way:

Step 1: Press the Calc menu ; Highlight the 'Probability Distributions'.

Step 2: Press $\underset{¯}{\text{Normal... ;}}$

Step 3: Tick $\square$ Cumulative Probability and enter the given data values

Mean: $0$

Standard deviation: $1$

Step 4: Tick $\square$ Input constant and enter the value $-0.633$

Input constant: $-0.633$

Step 5: Press Ok