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