Below is a graph of a normal distribution with mean --4 and standard deviation a-4. The shaded region represents the probability of obtaining a value from this distribution that is less than - 10. Shade the corresponding region under the standard normal density curve below.

Determine the area under the standard normal curve that lies to the right of (a) Z=-1.15, (b) Z=-1.69, (c) Z=0.45, and (d) Z=0.49. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) The area to the right of Z=-1.15 is 0.8749. (Round to four decimal places as needed.) (b) The area to the right of Z= -1.69 is (Round to four decimal places as needed.) Standard Normal Distribution Table (page 1) Area z -3.4 -3.3 -3.2 -3.1 -3.0 -29 -2.8 -2.7 -2.6 -2.5 -2.4 -2.3 -2.2 -2.1 -2.0 -1.9 -1.8 -1.7 -1.6 -1.5 -1.4 -1.3 -1.2 -1.1 -1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 -0.0 z 0.00 0.01 0.0003 0.0003 0.0005 0.0005 0.0007 0.0007 0.0006 0.0010 0.0009 0.0009 0.0013 0.0013 0.0019 0.0018 0.0026 0.0025 0.0035 0.0034 0.0047 0.0045 0.0062 0.0060 0.0082 0.0080 0.0107 0.0104 0.0139 0.0136 0.0179 0.0174 0.0228 0.0222 0.0287 0.0281 0.0359 0.0351 0.0446 0.0436 0.0548 0.0537 0.0668 0.0655 0.0808 0.0793 0.0968 0.0951…

Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean ? = 27.1 kilograms and standard deviation ? = 4.0 kilograms. Let x be the weight of a fawn in kilograms.

Find the​ Z-scores that separate the middle 22​% of the distribution from the area in the tails of the standard normal distribution. rounding to two decimals -, and + anwser .

Find the area of the shaded region under the standard normal distribution between the given z-scores -1.95 and 1.05.

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.