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Sagot :
The standard deviation of the sampling distribution is 0.03
Instructor number is calculated as - (failure /100)
Number of failures for instructor A = 13 = 0.13
Number of failures for instructor B = 0 = 0.
Number of failures for instructor C = 11 = 0.11
Number of failures for instructor D = 16 = 0.16
Calculating mean -
Mean(m)
= 0.13 + 0.11 + 0.16) / 3
= 0.4 / 3
= 0.13
Calculating the Standard deviation -
Σ(X - m)² / N - 1
Σ(0.13 - 0.1333)² + (0.11 × 0.1333)² + ( 0.16 - 0.1333)² = 0.001266
= 0.001266 / 2
= 0.00063343
= √0.00063343
= 0.025168
= 0.03
Complete question -
The chair of the operations management department at Quality University wants to construct a p-chart for determining whether the four faculty teaching the basic P/OM course are in control with regard to the number of students who fail the course. Accordingly, he sampled 100 final grades from last year for each instructor, with the following results: Instructor Number of Failures Prof. A 13 Prof. B 0 Prof. C 11 Prof. D 16 What is the estimate of the standard deviation of the sampling distribution for an instructor's sample proportion of failures?
Read more about standard deviation on:
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