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- 9.) The mean starting salary for teachers is $53,475 nationally. The standard deviation is approximately $5,250. Assume that the starting salary is normally distributed. a.) State the random variable. b.) Find the probability that a starting teacher will make more than $60,000. c.) Find the probability that a starting teacher will make less than $50,000. d.) Find the probability that a starting teacher will make between $52,000 and $59,500. e.) If a teacher made more than $61,000, would you think the teacher was overpaid? Why or why not?need help pleaseA waiterbelieves the distribution of his tips has a model that is slightly skewed to the left, with a mean of $8.60 and a standard deviation of $4.20. He usually waits on about 50 parties over a weekend of work. a) Estimate the probability that he will earn at least $450. b) How much does he earn on the best 5% of such weekends?
- Citizens use an average of 650 pounds of paper a year. Suppose that the distribution is normal with a population standard deviation of 153.5 pounds. Suppose you take a random sample of 25 Citizens. What is the probability that the mean pounds of paper used in a year are between 500 and 700 poundsSuppose that the distribution of the grades of students in MAPEH performance 1 is roughly normally distributed with a mean of 85 and a standard deviation of 5 a. Between what values do we expect the grades of the middle 68% of the MAPEH students in performance 1 lie? b. Suppose you randomly select 16 MAPEH students, find the probability that the average grades of these students is at least 86. c. A MAPEH student whose grades lies in the upper 1% of the distribution (inclusive of the 1% cutoff) is considered extremely performing in Performance 1. Can you say that Jazzer, a randomly selected student who obtained a grade of 95, is one of those who performed extremely in Performance 1?The average student loan debt for college graduates is $25,200. Suppose that that distribution is normal and that the standard deviation is $13,300. Let X = the student loan debt of a randomly selected college graduate. Round all probabilities to 4 decimal places and all dollar answers to the nearest dollar. a. What is the distribution of X? X - N b Find the probability that the college graduate has between $22,700 and $29,650 in student loan debt. c. The middle 30% of college graduates' loan debt lies between what two numbers? Low: $ High: S
- The probabilities that a customer buys 1, 2, 3, 5, 8 apples each day in a grocery store are 0.20, 0.32, 0.23, 0.18, 0.07 respectively. Let B be the number of apples a customer buys in a day. What is the standard deviation?e) Suppose that weekly household expenditures on petrol are normally distributed, with a mean of $120 and a standard deviation of $50. i. A family is randomly selected. Find the probability that this family spends more than $245 per week on petrol? A 2.5 B 0.0062 C 0.9938 D 0.0202 ii. The government wants to promote public transport and will award a prize to the families who spend the lowest amount on petrol. If the government only wants to give a prize to 10% of households, what is the most that a family is allowed to spend on petrol each week to obtain the prize? A $56 B $184 C $3.50 D $217.60 In a sample of 25 families, what is the probability that the average weekly expenditure on petrol will be less than $105? ii. A 0.0668 B 0.9332 C 0.3821 D 0.6179The average student loan debt for college graduates is $25,650. Suppose that that distribution is normal and that the standard deviation is $14,350. Let X = the student loan debt of a randomly selected college graduate. Round all probabilities to 4 decimal places and all dollar answers to the nearest dollar.b Find the probability that the college graduate has between $28,800 and $41,400 in student loan debt. c. The middle 20% of college graduates' loan debt lies between what two numbers? Low: $ High: $
- The median annual cost of auto insurance is $ 939 (According to DACO). Suppose the standard deviation is o = $ 241. What is the probability that in a simple random sample of auto insurance policies the sample mean differs more than $ 70.42 from the population mean if the sample size is 45? You must calculate the probability that the difference between the average cost in the sample and the average annual cost do not differ by $ 70.42. Find the probability that P (\ µ - š | 0) = P (-70.42 H-iS 70.42) Select one: a. The probability is 99% that the auto policy will not differ by $ 70.42 from the average annual cost. b. The probability is 50% that the auto policy will not differ by $ 70.42 from the average annual cost. c. The probability is 80% that the auto policy will not differ by $ 70.42 from the average annual cost. d. The probability is 70% that the auto policy will not differ by $ 70.42 from the average annual cost. e. The probability is 95% that the auto policy will not differ by $…Assume that the sales of Teri’s Specialty Cakes has historically followed a normal distribution, with an average of 250 cakes per month, and a standard deviation of 15 cakes.a. What is the probability that the sales for cakes will be no more than 285 over the next month?b. What is the probability that the sales for the cakes will be no more than 230 cakes over the next month?c. What is the probability that the sales for the cakes will be between 245 and 265 over the next month?d. What number of cake demand can Teri reasonably assume with a confidence level of 95%?For a, b, c, and d, please write out your work for at least one of these problems. For all problems, explain your answers and sketch a plot (similar to the handouts and text).
