Find the mean for the binomial distribution which has the stated values of n and p. n = 2142; p = 0.63
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n = 2142; p = 0.63
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- Assume that adults have IQ scores that are normally distributed with a mean of u= 100 and a standard deviation g= 15. Find the probability that a randomly selected adult has an IQ less than 118. Click to view page 1 of the table. Click to view page 2 of the table. The probability that a randomly selected adult has an IQ less than 118 is (Type an integer or decimal rounded to four decimal places as needed.)Solve for the mean and standard deviation of the following Binomial distributions. a. n = 20 and p = 0.70 b. n = 70 and p = 0.35 c. n = 100 and p = 0.50Find the z-score that has 70.54% of the distribution's area to its right. Click to view page 1 of the table. Click to view page 2 of the table. O A. - 0.82 В. - 0.54 С. 0.82 D. 0.54
- A researcher recorded the number of shots made, X, out of 10 free throws by a sample of 1000 randomly selected college students. The table below is a probability distribution table representing the data collected. Calculate the mean and the standard deviation of the probability distribution using a TI-83, TI-83 Plus, or TI-84 graphing calculator. x P(x) 0 0.01 1 0.01 2 0.02 3 0.04 4 0.03 5 0.05 6 0.21 7 0.29 8 0.21 9 0.08 10 0.05The average number of kilos of meat a person consumes in a year is 90 kilos. Assume that the standard deviation is 10 kilos and the distribution is approximately normal. If a sample of 45 individuals is selected, find the probability that the mean of the sample will be 87 kilos or greater per year. Express your answer in percent form with 2 decimal places (with percent symbol - e.g. 10.00%) type your answer...K Assume that adults have IQ scores that are normally distributed with a mean of μ = 105 and a standard deviation o=20. Find the probability that a randomly selected adult has an IQ between 89 and 121. Click to view page 1 of the table. Click to view page 2 of the table. The probability that a randomly selected adult has an IQ between 89 and 121 is (Type an integer or decimal rounded to four decimal places as needed.)
- The round off errors when measuring the distance that a long jumper has jumped is uniformly distributed between 0 and 5.8 mm. Round values to 4 decimal places when possible. b. The standard deviation is c. The probability that the round a. The mean of this distribution is off error for a jumper's distance is exactly 0.3 is P(x-03)-d. The probability that the round off error for the distance that a long jumper has jumped is between 0 and 5 8 mm is PC1.7 x 5.2)- that the jump's round off error is greater than 1.76 is P(x > 1.76) Find the 81st percentile. e. The probability f P(x > 1.4 x > 0.6) h. Find the mnmum for the upper quartile.Assume that a procedure yields a binomial distribution with n = 48 trials and the probability of success for one trial is p = 0.37. Find the mean for this binomial distribution. Round your answer to at least one decimal. μ = Find the standard deviation for this distribution. Round your answer to at least two decimals. σ =Assume that a procedure yields a binomial distribution with n = 513 trials and the probability of success for one trial is p = 0.27. Find the mean for this binomial distribution. (Round answer to one decimal place.) Find the standard deviation for this distribution. (Round answer to two decimal places.) O = Use the range rule of thumb to find the minimum usual value u-20 and the maximum usual value u+20. Enter answer as an interval using square-brackets only with whole numbers. usual values =
- The average number of kilos of meat a person consumes in a year is 100 kilos. Assume that the standard deviation is 11 kilos and the distribution is approximately normal. If a sample of 40 individuals is selected, find the probability that the mean of the sample will be 105 kilos or greater per year. Express your answer in percent form with 2 decimal places (with the percent symbol - e.g. 10.00%). type your answer... NextSuppose that the weight of an newborn fawn is Uniformly distributed between 2.2 and 3 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible. a. The mean of this distribution is b. The standard deviation is c. The probability that fawn will weigh exactly 2.8 kg is P(x = 2.8) = d. The probability that a newborn fawn will be weigh between 2.6 and 2.8 is P(2.6 < x < 2.8) =Find the z-scores for which 80% of the distribution's area lies between -z and z. 01.282 O -0.282 O 0.4 O 0.8
