Question

A car manufacturer takes an average of 17.5 hours to construct a car. This includes time for stamping, welding, painting, assembly and inspections. Construction times vary with a standard deviation of 30 minutes and these times follow a normal distribution.
a. Find the construction time of a car, which is on the 10th percentile of this distribution.
b. What is the probability that a randomly selected car manufactured at this plant takes between 18 and 19 hours to construct?
c. Find the probability that the construction time for a randomly selected car manufactured at this plant is less than 17 hours and 50 minutes.

Answer 1

Answer on Question #54355 – Math – Statistics and Probability

A car manufacturer takes an average of 17.5 hours to construct a car. This includes time for stamping, welding, painting, assembly and inspections. Construction times vary with a standard deviation of 30 minutes and these times follow a normal distribution.

a. Find the construction time of a car, which is on the 10th percentile of this distribution.

b. What is the probability that a randomly selected car manufactured at this plant takes between 18 and 19 hours to construct?

c. Find the probability that the construction time for a randomly selected car manufactured at this plant is less than 17 hours and 50 minutes.

Solution

Method 1

We need to find $a$ such that

$$P(X < a) = 0.1,$$

where $X \sim N\left(17.5; \left(\frac{30}{60}\right)^2\right)$ is a random normally distributed variable.

Using Microsoft Excel 2013 type

$$= NORM.INV(0,1; 17,5; 0,5)$$

and the answer is 16.859.

Method 2

We need to find $a$ such that

$$P(X < a) = 0.1,$$

where $X \sim N\left(17.5; \left(\frac{20}{60}\right)^2\right)$ is a random normally distributed variable.

Using statistical tables or the command

$$= NORM.S.INV(0,1)$$

in Microsoft Excel 2013 obtain that $10^{\text{th}}$ percentile of the standard normal variable is -1.282, $P(Z < -1.282) = 0.1$.

It is known that

$$Z = \frac {X - E (X)}{s d (X)},$$

where $X \sim N\left(17.5; \left(\frac{30}{60}\right)^2\right)$, $Z \sim N(0; 1)$ are two random normally distributed variables.

Given times follow a normal distribution with the average of $E(X) = 17.5$ hours and the standard deviation of $sd(X) = \frac{30}{60} = 0.5$ hour, equality

$$P (Z < - 1. 2 8 2) = 0. 1$$

is equivalent to

$$P \left(\frac {X - E (X)}{s d (X)} < - 1. 2 8 2\right) = 0. 1,$$

or

$$P (X < - 1. 2 8 2 s d (X) + E (X)) = 0. 1,$$

i.e.

$$\begin{array}{l} P (X < - 1. 2 8 2 \cdot 0. 5 + 1 7. 5) = 0. 1, \\ P (X < 1 6. 8 5 9) = 0. 1. \\ \end{array}$$

Thus, the construction time of a car, which is on the 10th percentile of this distribution, is 16.859 hours.

b. Method 1

We need to find probability

$$P (1 8 < X < 1 9),$$

where $X \sim N\left(17.5; \left(\frac{30}{60}\right)^2\right)$ is a random normally distributed variable.

Type

$$= \text{NORM.DIST}(19; 17, 5; 0, 5; \text{TRUE}) - \text{NORM.DIST}(18; 17, 5; 0, 5; \text{TRUE})$$

in Microsoft Excel 2013 and the answer is 0.1573.

Method 2

It is known that

$$Z = \frac {X - E (X)}{s d (X)},$$

where $X \sim N\left(17.5; \left(\frac{30}{60}\right)^2\right)$, $Z \sim N(0; 1)$ are two random normally distributed variables.

Given times follow a normal distribution with the average of $E(X) = 17.5$ hours and the standard deviation of $sd(X) = \frac{30}{60} = 0.5$ hour, the probability that a randomly selected car manufactured at this plant takes between 18 and 19 hours to construct is

$$\begin{array}{l} P (1 8 < X < 1 9) = P \left(\frac {1 8 - E (X)}{s d (X)} < \frac {X - E (X)}{s d (X)} < \frac {1 9 - E (X)}{s d (X)}\right) = P \left(\frac {1 8 - 1 7 . 5}{0 . 5} < Z < \frac {1 9 - 1 7 . 5}{0 . 5}\right) = \\ = P (1 < Z < 3) = P (Z < 3) - P (Z < 1). \\ \end{array}$$

From z-table or using

$$\begin{array}{l} = \text{NORM.S.DIST}(1; \text{TRUE}) \\ \end{array}$$

and

$$\begin{array}{l} = \text{NORM.S.DIST}(3; \text{TRUE}) \\ \end{array}$$

in Microsoft Excel 2013 we know

$$P(Z < 1) = 0.84135; \quad P(Z < 3) = 0.99865.$$

Thus,

$$P(18 < X < 19) = 0.9987 - 0.8413 = 0.1573.$$

c.

Method 1

We need to find probability $P\left(X < 17\frac{50}{60}\right)$, where $X \sim N\left(17.5; \left(\frac{30}{60}\right)^2\right)$ is a random normally distributed variable.

Type

$$= \text{NORM.DIST}(17 + 50/60; 17.5; 0.5; \text{TRUE})$$

in Microsoft Excel 2013 and the answer is 0.7475.

Method 2

It is known that

$$Z = \frac{X - E(X)}{sd(X)},$$

where $X \sim N\left(17.5; \left(\frac{30}{60}\right)^2\right)$, $Z \sim N(0; 1)$ are two random normally distributed variables.

Given times follow a normal distribution with the average of $E(X) = 17.5$ hours and the standard deviation of $sd(X) = \frac{30}{60} = 0.5$ hour, the probability that the construction time for a randomly selected car manufactured at this plant is less than 17 hours and 50 minutes will be

$$\begin{array}{l} P\left(X < 17\frac{50}{60}\right) = P\left(\frac{X - E(X)}{sd(X)} < \frac{17\frac{50}{60} - E(X)}{sd(X)}\right) = P\left(Z < \frac{17\frac{50}{60} - 17.5}{0.5}\right) = \\ = P\left(Z < \frac{17\frac{50}{60} - 17\frac{30}{60}}{0.5}\right) = P\left(Z < \frac{20}{60 \cdot 0.5}\right) = P\left(Z < \frac{20}{30}\right) = P(Z < 0.667) = 0.7475. \end{array}$$

From z-table we know

$$P(Z < 0.66) = 0.7454; \quad P(Z < 0.67) = 0.7486.$$

Type

$$= \text{NORM.S.DIST}(2/3; \text{TRUE})$$

in Microsoft Excel 2013 and the answer is 0.7475.

Answer:

a. 16.859;

b. 0.1573;

c. 0.7475.

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