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- What is the equation of the normal line to y=x at x=1? а. y-1%3-1/2(х-1) b. y-1=2(x-1) с. у-13-2(x-1)IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = 1Q of an individual. (a) Find the z-score for an IQ of 97, rounded to three decimal places. (b) Find the probability that the person has an IQ greater than 97. (c) Shade the area corresponding to this probability in the graph below. (Hint: The x-axis is the z- Score. Use your z-score from part (a), rounded to one decimal place). Shade: Left of a value Click and drag the arrows to adjust the values. -1 3 4 -1.5 (d) MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. (e) Sketch the graph, and write the probability statement. BIUX, x' C 次四 Edit. Insert - Formats - Σ ΣΗA researcher claims that the average wind speed in a certain city is 8 miles per hour. A sample of 32 days has an average wind speed of 8.2 miles per hour. The standard deviation of the population is 0.6 mile per hour. At 0.05, is there enough evidence to reject the claim? STEP 2. The critical value (s) is/are: z1 = z2 =
- Climate change and CO2 2011 Data collected fromaround the globe show that the earth is getting warmer.The most common theory relates climate change to anincrease in atmospheric levels of carbon dioxide (CO2),a greenhouse gas. The mean annual CO2 concentration inthe atmosphere (parts per million) is measured at the top of Mauna Loa in Hawaii, away from any local contami-nants. (Available at ftp://ftp.cmdl.noaa.gov/ccg/co2/ trends/co2_annmean_mlo.tx)The mean surface air temperature is recorded as thechange in °C relative to a base period of 1951 to 1980.(Available at data.giss.nasa.gov/gistemp/graphs_v3/)Here are a scatterplot and regression for the years from1959 to 2011:Dependent variable is TempR-squared = 82.3%s = 0.0985 with 53 - 2 = 51 degrees of freedomVariable Coefficient SE(Coeff) t-Ratio P-ValueIntercept -2.98861 0.2086 -14.3 ...0.0001CO2 0.0092 0.0006 15.4 ...0.0001a) Write the equation of the regression line.b) Is there evidence of an association between CO2 leveland global…a b Absolute standard deviation = Coefficient of variation = y = 1.20 (+0.02) x 10-8-3.60(±0.2) × 10-⁹ x Result = Absolute standard deviation = Coefficient of variation = C Result = d y=90.95 (±0.08) - 89.40(±0.06) +0.200(+0.004) Absolute standard deviation == Coefficient of variation = Result = ± y= y = 0.0040 (+0.0005) x 10.28 (+0.02) × 395(+1) Result = ± Absolute standard deviation = Coefficient of variation = % 329(+0.03) x 10-14 1.47 (+0.04) x 10-16 % % H % ±Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma.t Over a period of months, an adult male patient has taken nine blood tests for uric acid. The mean concentration was x = 5.41 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, with o = 1.75 mg/dl. (a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error? (Round your answers to two decimal places.) lower limit upper limit margin of error (b) What conditions are necessary for your calculations? (Select all that apply.) normal distribution of uric acid On is large O o is unknown O uniform distribution of uric acid O o is known (c) Interpret your results in the context of this problem. The probability that this interval contains the true average uric acid level for this patient is 0.95. The probability that this…
- The lengths of pregnancies in a small rural village are normally distributed with a mean of 261 days and a standard deviation of 17 days. What percentage of pregnancies last fewer than 314 days? P(x < 314 days) = P(z<IQ is normally distributed with a mean of 100 and a standard deviation of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual. (a) Find the z-score for an IQ of 99, rounded to three decimal places. (b) Find the probability that the person has an IQ greater than 99. (c) Shade the area corresponding to this probability in the graph below. (Hint: The x- axis is the z-score. Use your z-score from part (a), rounded to one decimal place). Shade: Left of a value -1 -1.5 0 +). Click and drag the arrows to adjust the values. 1 2 (d) MENSA is an organization whose members have the top 2% of all IQs. Find the minimum IQ needed to qualify for the MENSA organization. (e) Sketch the graph, and write the probability statement. Edit Insert Formats B I U x₂ x² A => EM e & N 18 (f) The middle 50% of IQs fall between what two values? (g) Sketch the graph and write the probability statement. Edit Insert Formats BIUX₂ X² A E = = E EC P R N • Σ+ Σ Α Σ+ Σ ΑOverproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma.† Over a period of months, an adult male patient has taken six blood tests for uric acid. The mean concentration was x = 5.35 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, with σ = 1.83 mg/dl. (a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error? (Round your answers to two decimal places.) lower limit= upper limit= margin of error= (d) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.02 for the mean concentration of uric acid in this patient's blood. (Round your answer up to the nearest whole number.)
- If X is normally distributed and P(X<20)=0.5 and P(X>26)=0.0228, find the mean and standard deviation of X. Give a detailed solutionSuppose X is a non-standard normal variable, with a mean of 190, and std deviation of 7 (i.e. X~N(190,7)). You have an observation (z) on a standard normal scale (i.e. Z~N(0,1)). If z = -1.15, what is it on an X scale? Express your answer to two decimal places (ex: 123.45).A random sample of 17 adult male wolves from the Canadian Northwest Territories gave an average weight x1 = 98.6 pounds with estimated sample standard deviation s1 = 5.6 pounds. Another sample of 26 adult male wolves from Alaska gave an average weight x2 = 88.2 pounds with estimated sample standard deviation s2 = 6.0 pounds. (a) Categorize the problem below according to parameter being estimated, proportion p, mean μ, difference of means μ1 – μ2, or difference of proportions p1 – p2. Then solve the problem. μp μ1 – μ2p1 – p2 (b) Let μ1 represent the population mean weight of adult male wolves from the Northwest Territories, and let μ2 represent the population mean weight of adult male wolves from Alaska. Find a 99% confidence interval for μ1 – μ2. (Use 1 decimal place.) lower limit upper limit (c) Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of…