....... Standard normal (z) distribution ........ POSITIVE z Scores 0 Z Cumulative Area from the LEFT N .00 .01 .02 03 04 05 06 .07 08 0.0 .5000 .5040 5080 .5120 5160 .5199 5239 .5279 .5319 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 0.2 .5793 .5832 .5871 5910 .5948 .5987 6026 6064 6103 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844Standard normal (z) distribution X 3.0 .9987 .9987 .9987 9988 9988 9989 9989 .9989 9990 3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 3.2 .9993 .9993 9994 .9994 9994 .9994 .9994 .9995 .9995 3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 9997 .9997 3.50 and up .9999 Z .00 .01 .02 .03 04 .05 06 .07 .08 NOTE: For values of z above 3.49, use 0.9999 for the area. Common Crit Confidence *Use these common values that result from interpolation: Level 0.90 z Score Area 0.95 1.645 0.9500 0.99 2.575 0.9950\fUse the sample data and confidence level given below to complete parts (a) through (d). A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n = 970 and x = 508 who said "yes." Use a 90% confidence level. Click the icon to view a table of z scores. . . . a) Find the best point estimate of the population proportion p. (Round to three decimal places as needed.)