The “cold start ignition time” of an automobile engine is

Chapter 6, Problem 69E

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QUESTION:

Problem 69E

The “cold start ignition time” of an automobile engine is being investigated by a gasoline manufacturer. The following times (in seconds) were obtained for a test vehicle: 1.75, 1.92, 2.62, 2.35, 3.09, 3.15, 2.53, 1.91.

(a) Calculate the sample mean, sample variance, and sample standard deviation.

(b) Construct a box plot of the data.

Questions & Answers

QUESTION:

Problem 69E

(a) Calculate the sample mean, sample variance, and sample standard deviation.

(b) Construct a box plot of the data.

ANSWER:

Answer

Step 1 of 2

(a)

The following times (in seconds) were obtained for a test vehicle:

$1.75,\ 1.92,\ 2.62,\ 2.35,\ 3.09,\ 3.15,\ 2.53,\ 1.91.$

We are asked to find the sample mean, sample variance and standard deviation.

If the $n$ observations in a sample are denoted by ${x}_{1},\ {x}_{2},\ {x}_{n},$ the sample mean is,

$\overline{x}=\frac{{x}_{1}+\ {x}_{2}+{..........+x}_{n}}{n}$ ……..(1)

Let $X$ be the cold start ignition time ( in seconds)

The number of measurements is ${n}_{X}=8.$

$\overline{x}=\frac{1.75+1.92+.........+2.53+1.91}{27}=2.415\ seconds$

The sample variance,

${s}^{2\ }=\frac{\sum\limits_{i}^{\ }({x}_{i}\ -\overline{x}{)}^{2}}{n\ -1}$ ……….(2)

We can rewrite equation (2) as,

${s}^{2\ }=\frac{\sum\limits_{\ }^{\ }{x}^{2}-\frac{{\left({\sum\limits_{\ }^{\ }x}\right)}^{2}}{n}}{n\ -1}$

[ $\sum\limits_{\ }^{\ }{x}^{2}=\ 48.6554,\ and\ \sum\limits_{\ }^{\ }x=19.32$ from the data]

${s}^{2\ }=\frac{48.6554-\frac{{\left({19.32}\right)}^{2}}{8}}{8\ -1}$

$\$ ${s}^{2}\ =0.285\ secon{d}^{2}$

${s}_{X}=\sqrt{{s}^{2}}=\sqrt{0.285}=0.534\ second$

Hence the sample mean, sample standard deviation and sample variance of the cold start ignition time are $2.415\ seconds,\ 0.534\ second,\ and\ 0.285\ secon{{d}^{\ }}^{2}$ respectively.