Name: Challenge #3 1. A simple random sample of 50 adults Is obtained and each person's red blood cell count is measured (in microliters). The sample mean is 4.63. The population standard deviation for red blood cell counts is .54. Construct a 99% confidence interval estimate of the mean red blood cell count of adults. Confidence interval CI = plus or minus z a/2 *(sd/sqrt(n)) X = mean CI = [4.63 plus or minus z a/2 (0.54/sqrt(54))] = [4.63-2.58*(0.07), 4.63 + 2.58*(0.07) = [4.44, 4.82]. Sd = standard deviation A = l - (confidence level/100) Za/2 = z table CI = confidence interval Mean (x) = 4.63 SD = 0.54 Size sample (n) = 54 2. What percentage of a normal distribution is within 2 standard deviations of the mean? (Select the closest answer.) a. b. c. d. 50% 95% 75% 100% Answer is \"b.\" 95% of the distribution lies within two standard deviations of the mean. 68% of the distribution lies with one standard deviation of the mean. 99.7% of the distribution lies within three standard deviations of the mean. This is called the empirical rule. 3. The area under the standard normal distribution between -2.5 and 1.5 is a. 0.1338 b. 0.1668 c. 0.4332 d. None of the above P(a less than or equal to z-b)=F(b)-F(a) P(x less than -2.5 = (-2.5-0)/1 -2.51=-2.5 P(z less than 2.5) from the standard normal table = P(x less than 2.5)=(2.5-0)/1 =2.5/1=2.5 P(z less than 2.5) from standard normal table = P(-2.5 less than x less than 2.5) = ________-__________ = 4. A real estate agent records the ages of 50 randomly selected home buyers in her sales area. The mean age is 38 years, with a sample standard deviation of 8 years. a. Find a 99% confidence interval for the population mean age. Confidence Interval Confidence Interval = CI = plus or minus t a/2 * (sd/sqt (n)) [.38 plus or minus ta/2 (8/sqrt(50)] X = mean [.38 - 2.68 *(1.414).38+2.68*(10414)] Sd = standard deviation = [3421, 41.79] A = l - (confidence level /100) CI= confidence interval Mean (x) = .38 Standard deviation (sd) = 8 Sample size (n) = 50 5. A simple random sample of 120 SAT scores has a mean of 1520. Assume that SAT scores have a standard deviation of 300. Construct a 95% confidence interval estimate of the mean SAT score. Confidence Interval confidence interval = CI = x plus or minus z a/2 * (sd/sqrt(n)) [1520 plus or minus z a/2 (300/sqrt(120))] X=mean [1520-196*(29.78), 1520+196*(29.78)] Sd= standard deviation = [1463.62, 1580.38] A = I - (confidence level/100) Z a/2 = z table value CI = Confidence interval Mean(x) = 1520 Sd = 300 Sample size (n) = 120