a) Pr{Y > 170} b) Pr{Y < 145} c) Pr{165< Y < 179}
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2) Assuming that for the height of women, μ = 162.1 cm and σ = 8.8 cm, find the following:
a) Pr{Y > 170}
b) Pr{Y < 145}
c) Pr{165< Y < 179}
Refer to problem (2). Give the following percentiles:
a) 97.7
b) 50
c) 33
d) 67
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- Use theorem 5.4.3 to form a proportion in which RS is a geometric mean. Hint RVSRST Exercises 1-6An agent for a property management company would like to be able to predict the monthly rental cost for apartments based on the size of the apartment as defined by square footage. A sample of the rent of 25 apartments in a college rental neighborhood was selected, and the information collected revealed the following: Apartment Size (Sq. Ft.) Monthly Rent ($) 1 850 950 2 1,450 1,600 3 1,085 1,200 4 1,232 1,500 5 718 950 6 1,485 1,700 7 1,136 1,650 8 726 935 9 700 875 10 956 1,150 11 1,100 1,400 12 1,285 1,650 13 1,985 2,300 14 1,369 1,800 15 1,175 1,400 16 1,225 1,450 17 1,245 1,100 18 1,259 1,700 19 1,150 1,200 20 896 1,150 21 1,361 1,600 22 1,040 1,650 23 755 1,200 24 1,000 800 25 1,200 1,750 e) Determine the coefficient of determination r2 and then completely interpret…An agent for a property management company would like to be able to predict the monthly rental cost for apartments based on the size of the apartment as defined by square footage. A sample of the rent of 25 apartments in a college rental neighborhood was selected, and the information collected revealed the following: Apartment Size (Sq. Ft.) Monthly Rent ($) 1 850 950 2 1,450 1,600 3 1,085 1,200 4 1,232 1,500 5 718 950 6 1,485 1,700 7 1,136 1,650 8 726 935 9 700 875 10 956 1,150 11 1,100 1,400 12 1,285 1,650 13 1,985 2,300 14 1,369 1,800 15 1,175 1,400 16 1,225 1,450 17 1,245 1,100 18 1,259 1,700 19 1,150 1,200 20 896 1,150 21 1,361 1,600 22 1,040 1,650 23 755 1,200 24 1,000 800 25 1,200 1,750 i) Determine a 95% interval estimate for the average rent of apartments with 1000…
- The table below summarizes data from a survey of a sample of women. Using a 0.01significance level, and assuming that the sample sizes of 800 men and 300 women are predetermined, test the claim that the proportions of agree/disagree responses are the same for subjects interviewed by men and the subjects interviewed by women. Does it appear that the gender of the interviewer affected the responses of women? Gender of Interviewer Man Woman Women who agree 498 247 Women who disagree 302 53 Compute the test statistic, rounding to three decimal places. Find the critical value(s). (Round to three decimal places) What is the conclusion based on the hypothesis test?Prior to assessment of the outcome, the researchers did a manipulation check. Members 1 and 2 rated the attractiveness of the person in the photo. They reported that for the attractive photo, M=7.53, for the unattractive photo, M=3.20,F(1, 108) =184.29.The Prevalence per 10,000 population of having health insurance for 1 (Table 5 box s. above) is: The Prevalence per 10,000 population of having health insurance for 2 (Table 5 box w. above) is:
- 2- An expert estimates that the distribution parameter for durability times of parts produced with machine A in the factory is different from the distribution parameter for durability times of parts produced with machine B. Durability times of 4 parts produced from machine A and 4 parts produced from machine B are given below. Find the Mann-Whitney U value by using these data. a) 18 B) 6 NS) 16 D) 20 TO) 12STATE: How heavy a load (in pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Given are data from students doing a laboratory exercise. 33,190 31,860 32,590 26,520 33,280 32,320 33,020 32,030 30,460 32,700 23,040 30,930 32,720 33,650 32,340 24,050 30,170 31,300 28,730 31,920 We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume that characteristics of materials vary Normally. Suppose that the strength of pieces of wood like these follows a Normal distribution with standard deviation 3000 pounds. PLAN: We will estimate ? by giving a 98% confidence interval. SOLVE: Find the sample mean ?¯ . (Enter your answer rounded to the nearest whole number.) ?¯= Give a 98% confidence interval, [???,ℎ??ℎ] , for the mean load required to pull the wood apart. (Enter your answers rounded to the nearest whole number.) ???=…Answer the following using the Boxplot attached. 1a) Which region (North or South) has the town with the largest mortality?b) Which region (North or South) has the town with the smallest mortality?c) In terms of the central tendency, which region has a higher mortality?d) Is the mortality more dispersed in the North or the South?
