Use the probability distribution to complete parts (a) and (b) below. The number of defects per 1000 machine parts inspected Defects 0 1 2 3 41 Probability 0.265 0.286 0.245 0.154 0.037 Complete the table for x-values from 0 to 3. Use μ = 1.5. P(x) X-μ (x-μ)² (x-μ)²p(x) 0.265 -1.5 2.250 0.596 0.286 -0.5 0.245 0.5 0.154 1.5 (Round to three decimal places as needed.) XOT23 0 1 www 0.013

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Use probability distribution
Use the probability distribution to complete parts (a) and (b) below.
The number of defects per 1000 machine parts inspected
Defects
0
1
2
5
3
4
Probability 0.265
0.286
0.245 0.154
0.037 0.013
Complete the table for x-values from 0 to 3. Use μ = 1.5.
XOT 23
P(x)
0.265
0.286
0.245
0.5
0.154
1.5
(Round to three decimal places as needed)
0
2
2
X-μ (x-μ)² (x-μ)²p(x)
- 1.5
2.250
0.596
-0.5
1
INF
Transcribed Image Text:Use the probability distribution to complete parts (a) and (b) below. The number of defects per 1000 machine parts inspected Defects 0 1 2 5 3 4 Probability 0.265 0.286 0.245 0.154 0.037 0.013 Complete the table for x-values from 0 to 3. Use μ = 1.5. XOT 23 P(x) 0.265 0.286 0.245 0.5 0.154 1.5 (Round to three decimal places as needed) 0 2 2 X-μ (x-μ)² (x-μ)²p(x) - 1.5 2.250 0.596 -0.5 1 INF
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Introduction

In the given scenario, the machines were inspected. They counted the number of defects per 1000 machine parts. The distribution showed the number of defects and their respective probabilities. The defects ranged from 0 to 3 and the sum of all probabilities is equal to one. The mean defect value is given which is 1.5 (μ=1.5). We need to complete the table for the given number of defects (X) using the discrete probability distribution. 

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