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- Areas under the Normal Curve .00 .01 .02 .01 .02 .03 .04 .05 .06 .07 .08 -3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 -3.4 -3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003 -3.3 -3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005 -3.2 -3.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 -3.1 -3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010 -3.0 -2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 -2.9 -2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 -2.8 -2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026 -2.7 -2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036 -2.6 -2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048 -2.5 -2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064 -2.4 -2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094…A certain brokerage house wants to estimate the mean daily return on a certain stock. A random sample of 12 days yields the following return percentages.−1.81, 1.54, 1.52, −2.58, −2.3, 0.97, 0.93, −1.06, 1.04, 0.2, −0.63, −2.75 Send data to calculator If we assume that the returns are normally distributed, find a 90% confidence interval for the mean daily return on this stock. Then find the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit: ? Upper limit: ?A simple random sample of size n is drawn. The sample mean, x, is found to be 18.2, and the sample standard deviation, s, is found to be 4.5. Click the icon to view the table of areas under the t-distribution. (a) Construct a 95% confidence interval about u if the sample size, n, is 35. Lower bound: ; Upper bound: (Use ascending order. Round to two decimal places as needed.)
- Areas under the Normal Curve .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.5040 0.5438 0.5832 0.5199 0.5596 0.5239 0,5636 0.6026 0.5279 0,5675 0.6064 0.0 0.5000 0.5080 0.5120 0.5319 0.5714 0.6103 0.5160 0.5359 0.0 0.5753 0.1 0.2 0.5398 0.5478 0.5517 0.5557 0.1 0.5793 0.5871 0.5910 0.5948 0.5987 0.6141 0.2 0.6331 0.6700 0.6368 0.6480 0.6517 0.6879 0.6179 0.6217 0.6255 0.6628 0.6293 0.6664 0.6406 0.6443 0.6808 0.3 0.4 0.3 0.4 0.6554 0.6591 0.6736 0.6772 0.6844 0.6985 0.7324 0.7642 0.7224 0.7549 0.7852 0.5 0.6915 0.6950 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.5 0.6 0.7257 0.7291 0.7357 0.7389 0.7422 0.7734 0.7454 0.7486 0.7794 0.8078 0.7517 0.7823 0.6 0.7 0.7580 0.7611 0.7673 0.7704 0.7764 0.7 0.7995 0.8264 0.7939 0.8 0.9 0.8051 0.8315 0.8133 0.8389 0.7881 0.7910 0.7967 0.8023 0.8106 0.8 0.8159 0.8186 0.8212 0.8238 0.8289 0.8340 0.8365 0.9 1.0 0.8413 0.8485 0.8554 0.8770 0.8962 0.8599 0.8810 0.8438 0.8461 0.8508 0.8531 0.8577 0.8621 1.0 1.1 0.8643 0.8665 0.8686 0.8708 0.8830 0.8729 0.8925…A certain brokerage house wants to estimate the mean daily return on a certain stock. A random sample of 16 days yields the following return percentages. -0.3, -1.16, -0.45, 2.98, –0.29, 0.44, 0.31, -2.59, 0.73, 1.93, 1.84, - 1.62, – 1.65, 2.61, 0.18, -2.25 Send data to calculator Send data to Excel If we assume that the returns are normally distributed, find a 90% confidence interval for the mean daily return on this stock. Then find the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.) Lower limit: II Upper limit:IAssume that adults have lQ scores that are normally distributed with a mean of µ = 105 and a standard deviation g = 15 Find the probability that a randomly selected adult has an IQ between 88 and 122. Click to view page 1 of the table Click to view page 2 of the table. ... The probability that a randomly selected adult has an IQ between 88 and 122 is (Type an integer or decimal rounded to four decimal places as needed.)
- Standard Normal Dist. Table 1 z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09-3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002-3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .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 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048-2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064-2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084-2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110-2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143-2.0 .0228 .0222 .0217 .0212…A certain brokerage house wants to estimate the mean daily return on a certain stock. A random sample of 12days yields the following return percentages 2.23, 1.76, 0.43, 0.99, −2.9, −0.35, 1.69, −0.1, −0.62, −1.48, −2.53, 1.15. If we assume that the returns are normally distributed, find a 95% confidence interval for the mean daily return on this stock. Give the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. Lower limit= Upper limit=Underlying price currently at 200 and follows a uniform distribution with mean of 200. You observed 80 strike PUT priced at $5.00. What is the implied MAD?
- Standard Normal Dist. Table 1 z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09-3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002-3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .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 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048-2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064-2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084-2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110-2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143-2.0 .0228 .0222 .0217 .0212…Standard Normal Dist. Table 1 z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09-3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0002-3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 .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 .0060 .0059 .0057 .0055 .0054 .0052 .0051 .0049 .0048-2.4 .0082 .0080 .0078 .0075 .0073 .0071 .0069 .0068 .0066 .0064-2.3 .0107 .0104 .0102 .0099 .0096 .0094 .0091 .0089 .0087 .0084-2.2 .0139 .0136 .0132 .0129 .0125 .0122 .0119 .0116 .0113 .0110-2.1 .0179 .0174 .0170 .0166 .0162 .0158 .0154 .0150 .0146 .0143-2.0 .0228 .0222 .0217 .0212…Find the value of z if the area under a standard normal curve (a) to the right of z is 0.3228; (b) to the left of z is 0.1271; (c) between 0 and z, with z> 0, is 0.4890; and (d) between -zand z, with z> 0, is 0.9500. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table.