Given a normal random variable X with mean 68 and variance 25, and a random sample of size n taken from the distribution, what sample size n is necessary in order that P(67.8sXs68.2)=0.97? Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. The necessary sample size is n = (Round up to the nearest whole number.)
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- Describe the standard normal distribution. What are its characteristics?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.area under the normal curve
- Determine the area under the standard normal curve that lies to the right of (a) Z= - 0.27, (b) Z= - 0.68, (c) Z = 0.11, and (d) Z= – 1.16. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Standard normal distribution kable (page 2) (a) The area to the right of Z= - 0.27 is (Round to four decimal places as needed.) Area Standard Nornal Distribution 0.05 0.00 0.01 0.02 0.03 0.04 0.06 0.07 0.08 0.09 0.5000 0.5398 0.5793 0.6179 0.6554 0.5040 0.5438 0.5832 0.5080 0.5478 0.5871 0.6255 0.6628 05120 0.5517 0.5910 0.6293 0.6664 0.5160 0.5557 0.5948 0.5199 0.5596 0.5987 0.5239 0.5636 0.6026 0.6406 0.6772 0.5279 0.5675 0.6064 0.6443 0.6808 0.5319 0.5714 0.6103 0.5359 0.5753 0.6141 0.0 0.1 0.2 0.3 0.4 0.6217 0.6591 0.6331 0.6700 0.6368 0.6736 0.6480 0.6844 0.6517 0.6879 0.7190 0.7517 0.7823 0.7224 0.5 0.6 0.7 0.6915 0.7257 0.7580 0.7881 0.8159 0.6950 0.7291 0.7611 0.7910 0.8186 0.6985 0.7324 0.7642 0.7019…Find the area to the left of z = -1.5 Solution: Draw the normal curve and shade the required region 2 3 Find the area that z = -1.5 corresponds to an area corresponds to the z-value Examine the graph and use probability notation to form the equation . The proportion of the area to the left of z = -1.5 is9. A simple random sample of size n is drawn. The sample mean, x, is found to be 18.3, and the sample standard deviation, s, is found to be 4.8. 15 Click the icon to view the table of areas under the t-distribution. (a) Construct a 95% confidence interval about µ if the sample size, n, is 35. ; Upper bound: (Use ascending order. Round to two decimal places as needed.) Lower bound: (b) Construct a 95% confidence interval about µ if the sample size, n, is 61. Lower bound: Upper bound: (Use ascending order. Round to two decimal places as needed.) How does increasing the sample size affect the margin of error, E? O A. The margin of error decreases. O B. The margin of error does not change. C. The margin of error increases. (c) Construct a 99% confidence interval about u if the sample size, n, is 35. ; Upper bound: (Use ascending order. Round to two decimal places as needed.) Lower bound: Compare the results to those obtained in part (a). How does increasing the level of confidence affect…
- = okil Determine the area under the standard normal curve that lies to the right of (a) Z= -0.63, (b) Z=0.93, (c) Z=0.31, and (d) Z= -0.72. 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= -0.63 is (Round to four decimal places as needed.) 150 O E 4 a de Clear all 73°F SunnyDetermine the area under the standard normal curve that lies between (a) Z=-1.55 and Z=1.55, (b) Z=-1.14 and Z=0, and (c) Z=0.13 and Z=2.22. Click the icon to view a table of areas under the normal curve. olve this P (a) The area that lies between Z=-1.55 and Z=1.55 is (Round to four decimal places as needed.) View an example 4x IA Get more help - F4 LDC 0 Tables of Areas under the Normal Curve Area F7 PrtScnfa F8 TABLE V 2 .00 .08 -3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002 -3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003 -3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005 -3.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007 -3.0 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 00011 00011 0.0010 0.0010 -2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014 -2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019 -2.7 0.0035 0.0034 0.0033 0.0032 0.0031…About ___________% of the area under the curve of the standard normal distribution is between z=−0.145z=-0.145 and z=0.145z=0.145 (or within 0.145 standard deviations of the mean). About __________% of the area under the curve of the standard normal distribution is outside the interval z=[−1.48,1.48]z=[-1.48,1.48] (or beyond 1.48 standard deviations of the mean). About________ % of the area under the curve of the standard normal distribution is outside the interval z=−1.82z=-1.82 and z=1.82z=1.82 (or more than 1.82 standard deviations from the mean).
- 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).1. A normal distribution of scores has a standard deviation of 10 and a mean of 50. Find the z-scores and its area under the normal curve corresponding to each of the following values: a. A score that is 20 points above the mean. b. A score that is 10 points below the mean.