In this situation what is the standard deviation of the sampleproportion of stud

Question

In this situation what is the standard deviation of the sampleproportion of students who have taken an APcourse?

In a sample of 100 students,what is the probability that over 32%of the sample has AP credit?

In a sample of 200 students whatis the probability that over 32% ofthe sample had AP credit?

An environmental research firm wants to estimate the proportion ofcars in Jackson County Missouri that would be considered SUV's.Using the county's vehicle registration records they randomlyselect 220 vehicles. From thatlist they determined that 92 ofthem were SUV's.

Note: Report youranswers to 4 decimal places.

Find the standard error of the sample proportion in thissituation.

From this data we see that we are 90%confident that the true proportion of SUVs in Jackson CountyMissouri is between and .

Explanation / Answer

probability that the students come to college with credit foran advance placement (AP) course = p = 0.25 probability that the students come to college without creditfor an advance placement (AP) course = q = 0.75 n = 100 In this situation what is the standard deviation of the sampleproportion of students who have taken an AP course = = = (npq) = (100 x 0.25 x 0.75) = 4.33 In a sample of 100 students, what is the probability thatover 32% of the sample has APcredit? standard error, e = / n = 4.33 / 100 =0.433 = np = 100 x 0.25 = 25 P>0.32 hence probability = 1-0.32 = 0.68 from z table, value of z corresponding to 0.68 is 0.47 hence (x-)/e >0.47 (x - 25) / 0.433 > 0.47 x >= 25.204 In a sample of 200 studentswhat is the probability that over 32%of the sample had AP credit? = npq = 6.124 standard error, e = / n = 6.124 / 200 =0.433 = np = 200 x 0.25 = 50 P>0.32 hence probability = 1-0.32 = 0.68 from z table, value of z corresponding to 0.68 is 0.47 hence (x-)/e >0.47 (x - 50) / 0.433 > 0.47 x >= 50.204 An environmentalresearch firm wants to estimate the proportion of cars in JacksonCounty Missouri that would be considered SUV's. Using the county'svehicle registration records they randomly select 220 vehicles. From that list they determinedthat 92 of them wereSUV's. hence p = 92 / 220 =0.418 q = 1-p = 0.582 n = 220 standard deviation, = npq = (220x0.418 x 0.582) =7.32 the standard error ofthe sample proportion in this situation , e = / n =7.32 / 220 = 0.494 From this data we seethat we are 90% confident that thetrue proportion of SUVs in Jackson County Missouri isbetween ?? = np = 220 x 0.418= 91.96 for probability = 0.9, zvalue = 1.29 hence confidenceinterval = ± ze = 91.96 ± (1.29 x 0.494) = ( 91.32274, 92.59726) In a sample of 200 studentswhat is the probability that over 32%of the sample had AP credit? = npq = 6.124 standard error, e = / n = 6.124 / 200 =0.433 = np = 200 x 0.25 = 50 P>0.32 hence probability = 1-0.32 = 0.68 from z table, value of z corresponding to 0.68 is 0.47 hence (x-)/e >0.47 (x - 50) / 0.433 > 0.47 x >= 50.204 An environmentalresearch firm wants to estimate the proportion of cars in JacksonCounty Missouri that would be considered SUV's. Using the county'svehicle registration records they randomly select 220 vehicles. From that list they determinedthat 92 of them wereSUV's. hence p = 92 / 220 =0.418 q = 1-p = 0.582 n = 220 standard deviation, = npq = (220x0.418 x 0.582) =7.32 the standard error ofthe sample proportion in this situation , e = / n =7.32 / 220 = 0.494 From this data we seethat we are 90% confident that thetrue proportion of SUVs in Jackson County Missouri isbetween ?? = np = 220 x 0.418= 91.96 for probability = 0.9, zvalue = 1.29 hence confidenceinterval = ± ze = 91.96 ± (1.29 x 0.494) = ( 91.32274, 92.59726) An environmentalresearch firm wants to estimate the proportion of cars in JacksonCounty Missouri that would be considered SUV's. Using the county'svehicle registration records they randomly select 220 vehicles. From that list they determinedthat 92 of them wereSUV's. hence p = 92 / 220 =0.418 q = 1-p = 0.582 n = 220 standard deviation, = npq = (220x0.418 x 0.582) =7.32 the standard error ofthe sample proportion in this situation , e = / n =7.32 / 220 = 0.494 From this data we seethat we are 90% confident that thetrue proportion of SUVs in Jackson County Missouri isbetween ?? = np = 220 x 0.418= 91.96 for probability = 0.9, zvalue = 1.29 hence confidenceinterval = ± ze = 91.96 ± (1.29 x 0.494) = ( 91.32274, 92.59726)

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