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Use the Standard Normal Table to find the corresponds to the cumulative area or

ID: 3217621 • Letter: U

Question

Use the Standard Normal Table to find the corresponds to the cumulative area or percentile. If the area is not in the table, use the entry closest to that area. If the area a halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. 0.2090 0.4364 0.9916 0.7995 0.05 0.85 0.94 0.0046 P_15 P_30 P_88 P_67 P_25 P_40 P_75 P_80 In Exercises 17-22, find the indicated z-score(s) shown in the graph. If convenient, use technology to find the z-score(s). In Exercises, 23-30, find the indicated z-score. Find the z-score that has 11.9% of the distribution's area to its left. Find the z-score that has 78.5% of the distribution's area to its left. Find the z-score that has 11.9% of the distribution's area to its right. Find the z-score that has 78.5% of the distribution s area to its right. Find the z-score for which 80% of the distribution s area lies between -z and z. Find the z-score for which 99% of the distribution s area lies between -z and z.

Explanation / Answer

2) use the standard normal table find the z score that corresponds to the cumulative area 0.4364

Ans:Using Excel=NORMSINV(0.4364)

                         =-0.16

4) use the standard normal table find the z score that corresponds to the cumulative area 0.7995

Ans:Using Excel=NORMSINV(0.7995)

                         =0.8398

24)find the z score that has 78.5% of the distrubution area to its left

Ans:Using Excel =NORMSINV(0.785)

                          =0.7891

24)find the z score that has 78.5% of the distrubution area to its right

Ans:=1-0.785

       =0.215

   Using Excel=NORMSINV(0.215)

                     =-0.7891

28)find the z score for which 99% of the distrubution area lies between -z and z

Ans:This would mean that the area from the left would be (1 - 0.99)/2 = 0.005. The only value of z which works in this case is -2.575, so z = 2.575 and -z = -2.575.

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