document.write( "Question 879203: 1.
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document.write( "Convert x = 70 to z-score if Normal Distribution has mean = 50 and Standard Deviation = 10.\r
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document.write( "A)
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document.write( "0.5
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document.write( "B)
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document.write( "1.0
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document.write( "C)
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document.write( "1.5
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document.write( "D)
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document.write( "2.0
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document.write( "Question 2
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document.write( "1.
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document.write( "Find x-value for z-score=1.5 if Normal Distribution has mean = 50 and Standard Deviation = 10.\r
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document.write( "A)
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document.write( "65
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document.write( "B)
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document.write( "60
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document.write( "C)
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document.write( "55
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document.write( "D)
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document.write( "45
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document.write( "Question 3
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document.write( "1.
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document.write( "Using the Appendix table for standard normal distribution\r
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document.write( "find the area under the curve to the left of z = -1.25\r
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document.write( "A)
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document.write( "0.1432
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document.write( "B)
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document.write( "0.3461
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document.write( "C)
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document.write( "0.4574
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document.write( "D)
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document.write( "0.1056
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document.write( "Question 4
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document.write( "1.
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document.write( "Use Appendix Table for standard normal distribution\r
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document.write( "to find area under the curve to the right of z = 2.00\r
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document.write( "A)
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document.write( "0.4572
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document.write( "B)
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document.write( "0.0228
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document.write( "C)
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document.write( "0.1462
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document.write( "D)
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document.write( "0.5431
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document.write( "Question 5
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document.write( "1.
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document.write( "Use standard normal distribution table (Appendix Table II) to find\r
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document.write( "the probability that z is between -1.5 and 1.5: P(-1.5 < z < 1.5)\r
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document.write( "A)
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document.write( "0.9332
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document.write( "B)
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document.write( "0.4268
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document.write( "C)
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document.write( "0.8664
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document.write( "D)
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document.write( "0.0559
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document.write( "Question 6
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document.write( "1.
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document.write( "For Normal distribution with mean 50 and standard deviation 10\r
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document.write( "find probability that x will be less than 35: P(x < 35).\r
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document.write( "A)
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document.write( "0.8893
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document.write( "B)
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document.write( "0.2101
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document.write( "C)
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document.write( "0.1020
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document.write( "D)
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document.write( "0.0668
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document.write( "Question 7
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document.write( "Household Income has Normal Distribution with average 44,000\r
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document.write( "and Standard Deviation = 20,000. Find the probability that\r
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document.write( "Household Income is GREATER than 56,000.\r
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document.write( "A)
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document.write( "0.1245
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document.write( "B)
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document.write( "0.2743
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document.write( "C)
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document.write( "0.4184
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document.write( "D)
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document.write( "0.6105
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document.write( "Question 8
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document.write( "Number of students who graduate College every year is Normal Distribution\r
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document.write( "with mean = 400 and standard deviation 50. Find probability that this year\r
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document.write( "between 300 and 500 students will graduate College.\r
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document.write( "Tip: Convert 300 and 500 to z values, use Appendix Table II\r
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document.write( "and then apply rule: P(a < x < b) = P(x < b) - P(x < a)\r
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document.write( "A)
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document.write( "0.52
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document.write( "B)
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document.write( "0.65
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document.write( "C)
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document.write( "0.95
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document.write( "D)
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document.write( "0.81
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document.write( "Question 9
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document.write( "1.
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document.write( "Sample size is 64, sample mean is 50.\r
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document.write( "Assuming the population standard deviation is ?=16,\r
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document.write( "construct 95% confidence interval for population mean.\r
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document.write( "For z factor use z?/2 = 2\r
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document.write( "A)
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document.write( "from 46 to 54
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document.write( "B)
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document.write( "from 40 to 60
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document.write( "C)
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document.write( "from 50 to 64
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document.write( "D)
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document.write( "from 35 to 65
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document.write( "1 points\r
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document.write( "Question 10
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document.write( "1.
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document.write( "For sample mean x = 80, sample size n = 36 and sample standard deviation s = 24\r
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document.write( "find 90% confidence interval for population mean.\r
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document.write( "This is the case when population standard deviation is not known.\r
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document.write( "Use formula on page 328 with t-value.\r
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document.write( "Find t-value in Appendix Table IV.\r
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document.write( "Tip: For confidence level 90% ? = 1 - 0.90 = 0.10, ?/2 = 0.10/2 = 0.05\r
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document.write( "So, in Table IV use column t0.05 and line with df = n - 1 = 36 - 1 = 35.\r
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document.write( "A)
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document.write( "from 65.48 to 95.86
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document.write( "B)
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document.write( "from 75.44 to 82.16
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document.write( "C)
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document.write( "from 73.24 to 86.76
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document.write( "D)
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document.write( "from 70.18 to 88.65
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document.write( "
Algebra.Com's Answer #530595 by ewatrrr(24785)![]() ![]() You can put this solution on YOUR website! Tell me do You have an TI Calculator? \n" ); document.write( "Are You checking Your answers? or not a clue? \n" ); document.write( "Question 1: z = 2 *Note: \n" ); document.write( "Question 2: x = 65 \n" ); document.write( "Question 3: P(z < -1.5)= .0668 \n" ); document.write( "Question 4: P(z > 2)= .0228 \n" ); document.write( "Question 5: P(-1.5 < z < 1.5) = .9332 -.0668 \n" ); document.write( "Question 6: P(x < 35) = P(z < -1.5)= .0668 *Note: \n" ); document.write( "\n" ); document.write( "Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted. \n" ); document.write( "The area under the normal curve to the left of a particular z-value = P(z) \n" ); document.write( "Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50% to the right \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |