Question

In this problem, you are going to numerically verify that the Central Limit Theorem is valid even when sampling from non-normal distributions.

Suppose that a component has a probability of failure described by a Weibull distribution.

Let X be the random variable that denotes time until failure; its probability density is:

f X(x; γ, k) = (k/γ)(x/γ)^k−1 e ^−(x/γ)k ,

for x ≥ 0, and zero elsewhere. In this problem, assume k = 2, γ = 125 [hours].

a) Simulate drawing a set of N = 20 sample values, repeated over M = 200 samples, from this component using the inverse sampling method. Plot the histogram of all N X M = 4,000 samples to verify that your samples represent the pdf above.

b) Compute the means associated with each of the M samples and plot a histogram of the sample means. Does this histogram appear to be normally distributed?

c) Using the histogram you obtained in a), estimate the probability that the component fails before 50 hours of operation.

Answer 1

Ans.solve this problem in Minitab

a)

MTB > Random 20 C1-C200;
SUBC> Weibull 2 125.
MTB > Histogram C1 - C200;
SUBC> Bar;
SUBC> Distribution;
SUBC> Normal;
SUBC> Overlay.

Histogram of C1, C2, C3, C4, C5, C6, C7, C8, ...

MTB > Describe C1 - C200;
SUBC> Mean.

Descriptive Statistics: C1, C2, C3, C4, C5, C6, C7, C8, ...

Variable Mean
C1 106.4
C2 90.8
C3 119.9
C4 117.1
C5 114.8
C6 105.3
C7 102.15
C8 105.7
C9 104.0
C10 91.5
C11 91.5
C12 107.6
C13 121.5
C14 94.7
C15 108.1
C16 104.7
C17 129.8
C18 118.6
C19 111.3
C20 92.2
C21 96.1
C22 87.5
C23 125.2
C24 106.0
C25 129.6
C26 110.8
C27 116.01
C28 95.0
C29 97.6
C30 114.7
C31 95.32
C32 127.4
C33 139.3
C34 127.0
C35 120.8
C36 108.41
C37 115.8
C38 86.0
C39 107.9
C40 111.0
C41 105.65
C42 118.6
C43 111.0
C44 122.3
C45 91.18
C46 118.8
C47 110.5
C48 105.0
C49 133.4
C50 121.1
C51 120.5
C52 104.5
C53 92.3
C54 117.0
C55 112.4
C56 121.4
C57 111.9
C58 123.2
C59 117.5
C60 106.8
C61 91.4
C62 90.3
C63 119.6
C64 105.67
C65 125.9
C66 114.3
C67 103.5
C68 102.87
C69 97.9
C70 101.6
C71 91.2
C72 104.7
C73 93.8
C74 116.6
C75 100.7
C76 110.5
C77 97.0
C78 133.2
C79 116.9
C80 126.3
C81 129.5
C82 118.9
C83 91.9
C84 114.5
C85 130.9
C86 102.0
C87 103.0
C88 113.6
C89 118.7
C90 126.9
C91 135.1
C92 121.5
C93 104.8
C94 87.0
C95 127.6
C96 98.9
C97 132.0
C98 85.6
C99 111.8
C100 104.8
C101 91.8
C102 124.6
C103 114.0
C104 97.6
C105 134.6
C106 113.5
C107 91.73
C108 111.2
C109 113.3
C110 110.4
C111 111.9
C112 101.9
C113 119.3
C114 89.2
C115 136.2
C116 124.2
C117 87.22
C118 114.5
C119 105.6
C120 142.2
C121 111.2
C122 119.8
C123 120.4
C124 104.6
C125 91.1
C126 108.29
C127 108.7
C128 89.4
C129 103.6
C130 92.8
C131 96.9
C132 118.1
C133 97.0
C134 111.1
C135 132.4
C136 117.4
C137 96.3
C138 103.2
C139 123.9
C140 101.7
C141 105.7
C142 116.6
C143 99.2
C144 111.2
C145 107.5
C146 121.6
C147 89.5
C148 93.13
C149 102.8
C150 94.5
C151 96.8
C152 91.9
C153 121.0
C154 96.7
C155 113.2
C156 95.9
C157 106.77
C158 111.1
C159 99.2
C160 124.8
C161 98.6
C162 130.6
C163 111.0
C164 109.3
C165 108.5
C166 118.3
C167 103.1
C168 121.7
C169 107.0
C170 116.1
C171 106.5
C172 105.1
C173 108.27
C174 123.8
C175 107.38
C176 108.3
C177 117.4
C178 109.7
C179 109.0
C180 97.0
C181 112.3
C182 138.7
C183 116.7
C184 97.1
C185 137.1
C186 94.4
C187 120.9
C188 88.16
C189 98.2
C190 117.0
C191 120.2
C192 115.0
C193 105.5
C194 112.7
C195 104.77
C196 95.0
C197 101.8
C198 98.3
C199 90.98
C200 118.4

MTB > Histogram 'Mean';
SUBC> Bar;
SUBC> Distribution;
SUBC> Normal;
SUBC> Overlay.

Histogram of Mean

a)

b)

The histogram of mean values is typically tending to look like a shape resembling the normal density.

c) it is calculated in excel

0.147856

=WEIBULL(50,2,125,TRUE)