# ch3_examples from mecha

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• Measurement in Statistics

Example: Find the Mean, Median, Standard Deviation, Variance and Mode

• Measurement in Statistics

Solution:

Mean

n

i

in

n

x

n

xxxx

1

21 ....

= 1103oC

Median

= 1104oC

Standard Deviation

= 5.79oC

2

1

( )

1

ni

i

x xs

n

Variance

= 33.49oC

22

1

( )

1

ni

i

x xs

n

Mode

= 1104oC

• Probability Density Function Example:

The life of a given ball bearing can be characterized by a probability distribution function, f(x) as shown in figure

(a) Calculate the expected life of the bearing

(b) If a bearing pick at random from this batch, what is the probability that its life (x) will be less than 20 h, greater than 20 h, and finally exactly 20 h?

• Probability Density Function

Solution:

(a)

(b)

• Normal (Gaussian) Distribution

Example:

The result of a test that follows a normal distribution have a mean value of 10.0 and a standard deviation of 1.

Find the probability that a single reading is a) Between 9 12

b) Less or equal to 9

c) Greater than 12

• Normal (Gaussian) Distribution

Solution:

= 10.0 and = 1

a)

z1 = (9 10)/1 = -1 and z2 = (12 10)/1 = 2

Part 1: z = -1 to 0, Part 2: z = 0 to 2

Note for Part 1, z = -1 to 0 equal to z = 0 to 1

• Normal (Gaussian) Distribution

P(0 z 1) = 0.3413 & P(0 z 2) = 0.4772

P(-1 z 2) = 0.3413 + 0.4772 = 0.8185

• Normal (Gaussian) Distribution

b)

z1 = - and z2 = (9 10)/1 = -1

P(- z 0) = 0.5 & P(0 z 1) = 0.3413

P(- z -1) = 0.5 0.3413 = 0.1587

• Normal (Gaussian) Distribution

c)

z1 = (12 10)/1 = 2 and z2 =

P(0 z 2) = 0.4772 & P(0 z ) = 0.5

P(2 z ) = 0.5 0.4772 = 0.0228

• Interval Estimation of the Population Mean

Example:

Based on 36 reading, the average resistance is 25 and the sample standard deviation is 0.5.

Determine the 90% confidence interval of the mean resistance of the batch.

• Interval Estimation of the Population Mean

Solution:

Based on 36 reading, the average resistance is 25 and the sample standard deviation is 0.5.

90%, 1 = 0.90 and = 0.10

Since n > 30, can use normal distribution

• Interval Estimation of the Population Mean

36 reading n > 30, can use normal distribution

90%, 1 = 0.90 and = 0.10

Area fall within z = 0 & equal to 0.5

Area between z = 0 & z/2 is 0.5 - /2 = 0.45

• Interval Estimation of the Population Mean

Area between z = 0 & z/2 is 0.5 - /2 = 0.45

Average resistance: 25 0.14

• Interval Estimation of the Population Mean

Example:

A manufacturer of VCR systems would like to estimate the mean failure time of a VCR brand with 95% confidence. 6 systems are tested to failure, and the following data (in hours of playing time) are obtained: 1250, 1320, 1542, 1464, 1275 and 1383. Estimate the population mean and the 95% confidence interval on the mean.

• Interval Estimation of the Population Mean

Solution:

95% confidence >> = 0.05, /2 = 0.025

Since n = 6 < 30, use t-distribution

• Interval Estimation of the Population Mean

t/2 = 0.025 = 2.571, = n 1 = 5

• Least-Squares Linear Fit

Example:

From the following data obtain:

a) y as a function of x

b) Standard error of estimate

Xi 1 2 3 4

Yi 2.1 3.9 6.1 7.9

• Least-Squares Linear Fit Solution:

a)

1.96 0.10Y x

xi yi xi yi xi2 yi

2

1 2.1 2.1 1 4.41

2 3.9 7.8 4 15.21

3 6.1 18.3 9 37.21

4 7.9 31.6 16 62.41

10 20 59.8 30 119.24

22 )(

))((

ii

iiii

xxn

yxyxna

96.1)10()30)(4(

)20)(10()8.59)(4(2

a

22

2

)(

))((

ii

iiiii

xxn

yxxyxb

10.0)10()30)(4(

)8.59)(10()20)(30(2

b

• b) Standard error of estimate:

Least-Squares Linear Fit

2

2

n

yxaybyS

iiii

xy

2

)8.59)(96.1()20)(1.0(24.119 xyS

13.0xyS

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