ind the z value that corresponds to the given area in the figure below 0.4505. Use The Standard Normal Distribution Table and enter the answer to 2 decimal places.
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Find the z value that corresponds to the given area in the figure below 0.4505. Use The Standard
P(Z < z ) = 0.4505
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Solved in 3 steps with 2 images
- Find the area of the shaded region. The graph depicts the standard normal distribution with mean 0 and standard deviation 1. Click to view page 1 of the table. Click to view page 2 of the table. Z= 0.89 The area of the shaded region is (Round to four decimal places as needed.)Use the standard normal distribution table to find the area. Right of z=0.84Use the standard normal table to find the z-score that corresponds to the cumulative area 0.9616. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. Click to view page 1 of the standard normal table. Click to view page 2 of the standard normal table. (Type an integer or decimal rounded to three decimal places as needed.) - X - X Standard normal table (page 1) Standard normal table (page 2) - 3.3 .0004 0005 .0003 .0004 .0004 .0004 .0004 .0004 .0005 .0005 .0005 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 - 3.2 .0005 .0005 .0006 .0006 .0006 .0006 .0006 .0007 .0007 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 3.1 .0007 .0007 .0008 .0008 .0008 .0008 .0009 .0009 .0009 .0010 0.4 .6554 .6591 .6628 .6700 .6736 .6772 .6808 .6844 .6879 .6664 - 3.0 – 2.9 .0010 .0010 .0011 .0011 .0011 .0012 .0012 .0013 .0013 .0013 .6915 .7019 .7088 .7422…
- Find the value to the left of the mean so that 99.13% of the area under the distribution curve lies to the right of it. Use The Standard Normal Distribution Table and enter the answer to 2 decimal places. z=Use the standard normal table to find the z-score that corresponds to the cumulative area 0.0129. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. Click to view page 1 of the standard normal table. Click to view page 2 of the standard normal table. (Type an integer or decipaall eaundad to thcee deeisel elese apeaded - X Standard normal table (page 2) Standard normal table (page 1) Standard Normal Distribution Standard Normal Distribution Critical Values Critical Values Level of Confidence c Level of Confidence c 0.80 1.28 0.80 1.28 0.90 1.645 Area 0.90 1.645 0.95 1.96 0.95 1.96 0.99 2.575 0.99 2.575 .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 .09 .08 .07 .06 05 .04 .03 .02 .01 .00 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 - 3.4 - 3.3 - 3.2 - 3.1 - 3.0 - 2.9 .0002 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 0.1 .5398 .5438 .5478 .5517…Use the standard normal table to find the area to the left of z=−0.34. The area to the left of z=−0.34 is
- Use the standard normal table to find the z-score that corresponds to the cumulative area 0.9945 If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. Click to view page 1 of the standard normal table. Click to view page 2 of the standard normal table. (Type an integer or decimal rounded to three decimal places as needed.)The mean height of women in a country (ages 20 - 29) is 64.3 inches. A random sample of 70 women in this age group is selected. What is the probability that the mean height for the sample is greater than 65 inches? Assume o = 2.86. The probability that the mean height for the sample is greater than 65 inches is (Round to four decimal places as needed.)Please help
