Answer :
Answer:
Prof B and Prof D
Step-by-step explanation:
Step-by-step explanation:
From the question we are told that
Sample size 100
Instructor Number of Failures
Prof. A 13
Prof. B 0
Prof. C 11
Prof. D 16
Confidence level= 0.95
From Z table
Z=1.96
Generally proportion for failure is mathematically given as
[tex]Proportion\ of\ Failure\ P=\frac{Number\ of\ Failure\ N}{Sample\ size\ S}[/tex]
[tex]P=\frac{N}{S}[/tex]
For Prof A
[tex]P_A=\frac{N_A}{S}[/tex]
[tex]P_A=\frac{13}{100}[/tex]
[tex]P_A=0.13[/tex]
For Prof B
[tex]P_B=\frac{N_B}{S}[/tex]
[tex]P_B=\frac{0}{100}[/tex]
[tex]P_B=0[/tex]
For Prof C
[tex]P_C=\frac{N_C}{S}[/tex]
[tex]P_C=\frac{11}{100}[/tex]
[tex]P_C=0.11[/tex]
For Prof D
[tex]P_D=\frac{N_D}{S}[/tex]
[tex]P_D=\frac{16}{100}[/tex]
[tex]P_D=0.16[/tex]
Generally the average proportional failure is mathematically given as
[tex]P_a_v_g=\frac{0.13+0+0.11+0.16}{4}[/tex]
[tex]P_a_v_g=0.10[/tex]
Therefore having this we calculate for the control limits
Generally the upper control limit UCl is mathematically given as
Upper control limits:
[tex]UCL =P_a_v_g +Z\sqrt{\frac{P_a_v_g(1-P_a_v_g) }{S} }[/tex]
[tex]UCL =0.1 +1.96\sqrt{\frac{0.1(1-0.1) }{100}}[/tex]
[tex]UCL =0.1 +1.96*0.03[/tex]
[tex]UCL =0.1 +0.0588[/tex]
[tex]UCL =0.1588[/tex]
Generally the upper control limit UCl is mathematically given as
Lower limit control limits:
[tex]UCL =P_a_v_g _Z\sqrt{\frac{P_a_v_g(1-P_a_v_g) }{S} }[/tex]
[tex]UCL =0.1 -1.96\sqrt{\frac{0.1(1-0.1) }{100}}[/tex]
[tex]UCL =0.1 -1.96*0.03[/tex]
[tex]UCL =0.1 -0.0588[/tex]
[tex]UCL =0.0412[/tex]
Therefore the Range is 0.0412 to 0.1588
Given the range 0.0412 to 0.1588 Prof B and Prof D are outside the Range making them the correct options