-3 -2 -1 0 1 2 3 Please show your answer to 2 decimal places. About % of the area under the curve of the standard normal distribution (u = 0 and o = 1) is outside the interval Z = (-0.74,0.74) (or beyond 0.74 standard deviations of the mean).
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Given that
Z ~ N ( 0,1 )
About --- % Outside the interval z =( -0.74 ,0.74 )
Step by step
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- Next About % of the area under the curve of the standard normal distribution is outside the interval z = [-1.75, 1.75 (or beyond 1.75 standard deviations of the mean). Submit Question esc 80 F1 F2 F3 FA FS F6 @ %23 %24 2 3-3 -2 -1 0 1 2 3 Please show your answer to 2 decimal places. About % of the area under the curve of the standard normal distribution (u = 0 and o = 1) is outside the interval Z = (-1.02,1.02) (or beyond 1.02 standard deviations of the mean).After I round 15.15% to four decimal places = 0.1515 how do I start to find the mean?
- Find a Z value such that 90% of the standard normal curve life between negative Z and Z (round your answer to two decimal places) Z=About ______% of the area under the curve of the standard normal distribution is between z=−1.904 and z=1.904 (or within 1.904 standard deviations of the mean).Round your percent to two decimal placesFind the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15.
- -3 -1 0 1 -2 2 3 Please show your answer to 2 decimal places. About % of the area under the curve of the standard normal distribution (u = 0 and o = 1) is outside the interval Z = (-0.51,0.51) (or beyond 0.51 standard deviations of the mean).About % of the area under the curve of the standard normal distribution is outside the interval Z = (-1.07,1.07) (or beyond 1.07 standard deviations of the mean). Please show your answer to 2 decimal places.About _________% of the area under the curve of the standard normal distribution is between z=−0.928z=-0.928 and z=0.928z=0.928 (or within 0.928 standard deviations of the mean).
- Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. Click to view page 1 of th - × The area of the shaded r Standard Normal Table (Page 1) NEGATIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 02 .03 .04 .05 .06 .07 .08 .09 -3.50 and lower .0001 -3.4 .0003 .0003 .0003 .0003 .0003 .0003 -3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0003 0004 .0003 .0004 .0003 .0004 .0002 .0003 -3.2 .0007 .0007 .0006 0006 .0006 .0006 .0006 .0005 .0005 .0005 -3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 0007 .0007 -3.0 .0013 .0013 0013 .0012 0012 .0011 .0011 .0011 0010 0010 -2.9 .0019 0018 0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014 -2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 -2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026 -2.6 .0047 .0045 0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036 -2.5 .0062…Below is a graph of a normal distribution with mean u = -3 and standard deviation o =2. The shaded region represents theprobability of obtaining a value from this distribution that is between -4 and 0. Shade the corresponding region under the standard normal curve below.