I need them both please The average grade in a statistics course has been 76 wit

Question

I need them both please

The average grade in a statistics course has been 76 with a standard deviation of 11. If a random sample of 56 is selected from this population, what is the probability that the average grade is more than 80? Use Appendix B.1 for the z-values. (Round your z-value to 2 decimal places and the final answer to 4 decimal places.)

There are 440 welders employed at the Weller Shipyards Corporation. A sample of 50 welders revealed that 35 had graduated from a registered welding course. Construct the 99% confidence interval for the proportion of all welders who graduated from a registered welding course. (Round the final answers to 3 decimal places.)

Explanation / Answer

1) std error of mean =std deviation/(n)1/2 =11/(56)1/2 =1.47

therefore probability that the average grade is more than 80 =P(X>80)

=1-P(X<80)=1-P(Z<(80-76)/1.47)=1-P(Z<2.72)=1-0.9967=0.0033

2) here estimated proportion p=35/50=0.7

hence std error =((p(1-p)/n)*(N-n)/(N-1))1/2  where n=50 and N=440

=0.0611

for 99% confidence level; z=2.5758

therefore  99% confidence interval for the proportion of all welders who graduated from a registered welding course

= p -/+ z*std error =0.543 ; 0.857

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