Given a normal distribution with p = 103 and (I: 20, and given you select a sample of n = 16. complete parts (a) through (d). a. What is the probability that R is less than 90'? PUT 103.8) = 04364 (Type an integer or decimal rounded to four decimal places as needed.) d. There is a T031: chance thati is above what value? i: 0.03 (Type an integer or decimal rounded to two decimal places as needed.) Given a normal distribution with p 2 5i] and 0 = 4, and given you select a sample of r1 2100, complete parts (a) through (d) a. What is the probability that X is less than 219'? Fri <: an integer or decimal rounded to tour places as needed. b. what is the probability that i between and p919 r x above p> 50.5) = 0.1056 (Type an integer or decimal rounded to tour decimal places as needed.) d. There is a 35% chance thati is above what value? i = (Type an integer or decimal rounded to two decimal places as needed.) The weight of an energy bar is approximately normally distributed with a mean of 42.?5 grams with a standard deviation of l] 025 gram. Complete parts (a) through (e) below. a. What is the probability that an individual energy barweighs less than 42.735 grams? 0.27-1- {Round to three decimal places as needed.) b. If a sample of 4 energy bars is selected, what is the probability that the sample mean weight is less than 42.735 grams? 0.115 (Round to three decimal places as needed.) c. It a sample of 25 energy bars is selected, what is the probability that the sample mean weight is less than :12 T35 grams? 0.001 (Round to three decimal places as needed.) d. Explain the dilterence in the results at (a) and (c). Part (a) refers to an individual bar, which can be thought. of as a sample with sample size 5 . Therefore, the standard enor of the mean for an individual bar is 0.005 times the standard error of the sample in (c) with sample size 25. This leads to a probability in part. (a) that is larger than the probability in part (c). (Type integers or decimals. Do not round.) According to a report from a business intelligence company, smartphone owners are using an average of 19 apps per month. Assume that number of apps used per month by smartphone owners is normally distributed and that the standard deviation is 5. Complete parts (a) through (d) below. a. If you select a random sample of 36 smartphone owners, what is the probability that the sample mean is between 18.5 and 19.5? 0.452 (Round to three decimal places as needed.) b. If you select a random sample of 36 smartphone owners, what is the probability that the sample mean is between 18 and 19? 0.385 (Round to three decimal places as needed.) c. If you select a random sample of 144 smartphone owners, what is the probability that the sample mean is between 18.5 and 19.5? (Round to three decimal places as needed.)Full-time college students report spending a mean of 25 hours per week on academic activities, both inside and outside the classroom. Assume the standard deviation of time spent on academic activities is 5 hours. Complete parts (a) through [d] below. a. It you select a random sample of1Btulltime college students, what is the probability that the mean time spent on academic actiy'ties is at least 24 hours per week? 0.7531 (Round to four decimal places as needed ) b. If you select a random sample of 16 fulltime college students, there is an 31% chance that the sample mean is less than how manyr hours per week? 26.10 (Round to two decimal places as needed.) c. What assumption must you make in order to solve (a) and (b)? -:-'_._'A. The population is symmetrically distributed, such that the Central leil Theorem will llkely hold for samples of size 16. . The sample is symmetrically distributed, such that the Central Limit Theorem will likely hold. . The population is normally distributed. -: :1 D. The population is uniformly distributed. d. If you select a random sample of64 fulltime college students, there is an 31% chance that the sample mean is less than how many hours per week? (Round to two decimal places as needed.) A political pollster is conducting an analysis of sample results in order to make predictions on election night. Assuming a twtrcandidate election. if a specic candidate receives at least 55% of the vote in the sample, that candidate will be forecast as the winner of the election. You select a random sample of100 voters. Complete parts (a) through (c) below. a. What is the probability that a candidate will he forecast as the winner when the population percentage of her vote is 50.1 Ulla? The probability is 0.1635 that a candidate will be forecast as the winner when the population percentage of her vote is 50.1 91: (Round to four decimal places as needed.) b. What is the probability that a candidate will be forecast as the winner when the population percentage of her vote is 59%? The probability is 0.2090 that a candidate will be forecast as the winner when the population percentage other vote is 59%. (Round to four decimal places as needed.) A marketing survey is conducted in which students are to taste two different brands of soft drink. Their task is to correctly identify the brand tasted. A random sample of 180 students is taken. Assume that the students have no ability to distinguish between the two brands. Complete (a) through (d) below. a. What is the probability that the sample will have between 50% and 55% of the identifications correct? 0.4101 (Round to four decimal places as needed.) b. The probability is 90% that the sample percentage is contained within what symmetrical limits of the population percentage? Identify the limits of the population percentage. The lower limit is 0.4387 The upper limit is 0.5613 (Round to four decimal places as needed.) There is a 90% probability that the sample percentage will be contained within |% symmetrically around the population percentage. (Round to the one decimal place as needed.)