Statistics Questions & Answers

The sample size \( n \), probability of success \( p \), and probability of failure \( q \) are given for a binomial experiment. Decide whether you can use a normal distribution to approximate the distribution of \( x \). \( n=20, p=0.89, q=0.11 \) Can a normal distribution be used to approximate the distribution of \( x \) ? A. No, because \( n q<5 \). B. No, because \( n p<5 \) and \( n q<5 \). C. No, because \( n p<5 \). D. Yes, because \( n p \geq 5 \) and \( n q \geq 5 \).

In a survey of 18 -year-old males, the mean weight was 170.3 pounds with a standard deviation of 46.2 pounds. Assume the distribution can be approximated by a normal distribution. (a) What weight represents the 95th percentile? (b) What weight represents the 32nd percentile? (c) What weight represents the first quartile? (a) \( \square \) pounds (Round to one decimal place as needed.) (b) \( \square \) pounds. (Round to one decimal place as needed.) (c) \( \square \) pounds. (Round to one decimal place as needed.) (Rs

Question 2 Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 255 feet and a standard deviation of 57 feet. Use your graphing calculator to answer the following questions. Write your answers in percent form. Round your answers to the nearest tenth of a percent. a) If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled fewer than 222 feet? P(fewer than 222 feet) \( = \) b) If one fly ball is randomly chosen from this distribution, what is the probability that this ball traveled more than 234 feet?

The weights of ice cream cartons are normally distributed with a mean weight of 9 ounces and a standard deviation of 0.4 ounce. (a) What is the probability that a randomly selected carton has a weight greater than 9.15 ounces? (b) A sample of 16 cartons is randomly selected. What is the probability that their mean weight is greater than 9.15 ounces? (a) The probability is \( \square \). (Round to four decimal places as needed.) (b) The probability is \( \square \). (Round to four decimal places as needed.)

Chapter 12: The Normal Distribution - Homework Score: \( 0 / 100 \) Answered: \( 0 / 24 \) The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.9 days and standard deviation of 1.6 days. Use your graphing calculator to answer the following questions. Write your answers in percent form. Round your answers to the nearest tenth of a percent. a) What is the probability of spending less than 8 days in recovery? b) What is the probability of spending more than 5 days in recovery? c) What is the probability of spending between 5 days and 8 days in recovery? Question Help: Submit Question Video

A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1543 and the standard deviation was 319 . The test scores of four students selected at random are 1990, 1260, 2240, and 1430. Find the \( z \)-scores that correspond to each value and determine whether any of the values are unusual. The \( z \)-score for 1990 is (Round to two decimal places as needed.) The \( z \)-score for 1260 is (Round to two decimal places as needed.) The \( z \)-score for 2240 is (Round to two decimal places as needed.) The \( z \)-score for 1430 is (Round to two decimal places as needed.) Which values, if any, are unusual? Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The unusual value(s) is/are (Use a comma to separate answers as needed.) B. None of the values are unusual.

A convenience sample differs from a voluntary sample in that a convenience sample is structured based on accessibility to the researcher, and a voluntary sample is based on participant interest. convenience samples survey each participant once, and voluntary samples survey each participant numerous times. convenience sampling is a method of random sampling, and a voluntary sample is not. convenience sampling is not a probability-based method, and voluntary sampling is.

A drug tester claims that a drug cures a rare skin disease \( 74 \% \) of the time. The claim is checked by testing the drug on 100 patients. If at least 69 patients are cured, the claim will be accepted. Find the probability that the claim will be rejected assuming that the manufacturer's claim is true. Use the normal distribution to approximate the binomial distribution if possible. The probability is

In the population of Betta Imbellis (wild caught bettas, the Siamese Fighting Fish). Let \( p \) be the true proportion of those fishes that carry the green pigments in their eyes. A) In a random sample of 102 bettas, 30 of them carry the green pigments in their eyes. Construct a(n) \( 99 \% \) confidence interval to estimate \( p \). \( \mathrm{z}_{\mathrm{c}}=2.5758 . \quad \) (Round to 4 decimal places) The interval: \( 0.2941 \pm \square \) (Round \( E \) to 4 decimal places) The interval in traditional format:

B) In a random sample of 229 bettas, 58 of them carry the green pigments in their eyes. Construct a(n) \( 99 \% \) confidence interval to estimate \( p \). \( z_{c}=2.5758 \quad \) (Round to 4 decimal places) The interval: \( 0.2533 \pm \square \) (Round \( E \) to 4 decimal places) The interval in traditional format: ( )