What are the four conditions necessary to have a binomial distribution?43 2.43 Suppose the random variable X has a binomial distribution with n = 10 trials and probability of success p = 0.25. Using the probabilities given in Table 2.17, determineTable 2.17 The Binomial Probabilities for n = 10 Trials and p = 0.25Binomial with n = 10 and p = 0.25xP(X = x)00.05631410.18771220.28156830.25028240.14599850.05839960.01622270.00309080.00038690.000029100.000001the most likely value of X.the least likely value of X.the probability that X is less than 6.the probability that X is greater than equal to 4.the probability that 2≤X≤6.the mean value of X.
44 2.44 Determine the mean, variance, and standard deviation for each of the following binomial distributions.n = 50 and p = 0.4.n = 200 and p = 0.75.n = 80 and p = 0.25.
45 2.45 For what values of p will a binomial distributionhave a long tail to the right?have a long tail to the left?be symmetric?have the largest value of σ?
46 2.46 Many studies investigating extrasensory perception (ESP) have been conducted. A typical ESP study is carried out by subjecting an individual claiming to have ESP to a series of trials and recording the number of correct identifications made by the subject. Furthermore, when a subject is strictly guessing on each trial, the number of correct identifications can be modeled with a binomial probability model with the probability of a correct identification being p = 0.5 on each trial. If a subject is guessing on each of 20 trials in an ESP study, determinethe probability of 20 correct identifications.the probability of 18 correct identifications.the probability of at least 18 correct identifications.the mean number of correct identifications.
47 2.47 Suppose an individual actually does have ESP and makes correct identifications with probability p = 0.95. If the individual is subjected to a series of 20 independent trials, determinethe probability of making 20 correct identifications.the probability of making fewer than 19 correct identifications.the mean number of correct identifications.
48 2.48 Past studies have shown that 60% of the children of parents who both smoke cigarettes will also end up smoking cigarettes, and only 20% of children whose parents do not smoke cigarettes will end up smoking cigarettes. In a family with four children, use the binomial probability model to determinethe probability that none of the children become smokers given that both parents are smokers.the probability that none of the children become smokers given that none of the parents are smokers.
49 2.49 In Exercise 2.48, is it reasonable to assume that each of the four children will or will not become a smoker independently of the other children? Explain.
50 2.50 Side effects are often encountered by patients receiving a placebo in a clinical trial. Suppose 10 individuals were randomly and independently selected for the placebo group in a clinical trial. From past studies, it is known that the percentage of individuals experiencing significant side effects after receiving the placebo is about 10%. Using the binomial probability model, determinethe probability that two of the 10 patients in the placebo group experience significant side effects.the probability that none of the 10 patients in the placebo group experience significant side effects.the expected number of the 10 patients in the placebo group that will experience significant side effects.the standard deviation of the number of the 10 patients in the placebo group that will experience significant side effects.
51 2.51 If Z has a standard normal distribution, determineP(Z≤−0.76).P(Z<1.28).P(Z≤−2.04).P(Z>0.42).P(Z≥−1.65).P(Z>2.87).P(−1.12<Z≤2.25).P(1.10<Z<2.25).P(−0.80≤Z≤1.22).P(−1.76<Z<−1.26).
52 2.52 If Z has a standard normal distribution, determinethe 5th percentile.the 25th percentile.the 75th percentile.the 98th percentile.the interquartile range.
53 2.53 Intelligence quotient scores are known to follow a normal distribution with mean 100 and standard deviation 15. Using the normal probability model, determinethe probability that an individual has an IQ score of greater than 140.the probability that an individual has an IQ score of less than 80.the probability that an individual has an IQ score of between 105 and 125.the 95th percentile of IQ scores.
54 2.54 Suppose the birth weight of a full-term baby born in the United States follows a normal distribution with mean 7.5 pounds and standard deviation 0.5 pounds. Determine theprobability that a full-term baby born in the United States weighs between 7 and 8 pounds.probability that a full-term baby born in the United States weighs more than 9 pounds.probability that a full-term baby born in the United States weighs less than 6.8 pounds.5th percentile of the weights of full-term babies born in the United States.95th percentile of the weights of full-term babies born in the United States.
55 2.55 According to the National Health Statistics Report Number 122, December 20, 2018 (Fryar, 2018), the estimated mean weight of an adult male in the United States is 197.8 pounds. Suppose the distribution of weights of adult males in the US is normally distributed with mean weight µ = 200 pounds with standard deviation of σ = 25 pounds. Determine theprobability that an adult male in the US weighs more than 240 pounds.probability that an adult male in the US weighs less than 140 pounds.probability that an adult male in the US weighs between 180 and 220 pounds.90th percentile of the weights of adult males in the US.
56 2.56 According to the National Health Statistics Report Number 122, December 20, 2018 (Fryar, 2018), the estimated mean body mass index (BMI) of an adult female in the United States is 29.6. Suppose the distribution of BMI values for adult females in the US is normally distributed with mean BMI µ = 30 with standard deviation of σ = 4. Determine theprobability that an adult female in the US has BMI less than 25.probability that an adult female in the US has BMI more than 36.probability that an adult female in the US has BMI between 26 and 32.10th percentile of the BMI values of adult females in the US.
57 2.57 What are the units of a z score?
58 2.58 How many standard deviations below the mean does a z score of −3 correspond to?
59 2.59 Under what conditions is it possible to determine the percentile associated with an observed z score?
60 2.60 Table 2.18 contains the standard weight classifications based on body mass index (BMI) values. Assuming that BMI is approximately normally distributed, determine the z score corresponding to the cutoff for theTable 2.18 The Standard Weight Classifications Based on BMI ScoresWeight ClassificationBMI Percentile RangeUnderweightLess than 5th percentileHealthy weightBetween 5th and 85th percentilesAt risk of overweightBetween 85th and 95th percentilesOverweightGreater than the 95th percentileunderweight classification.healthy classification.at-risk-of-overweight classification.overweight classification.
61 2.61 Because a BMI value for a child depends on age and sex of the child, z scores are often used to compare children of different ages or sexes. Table 2.19 gives the mean and standard deviation for the distribution of BMI values for male children aged 10 and 15. Use the information in Table 2.19 to answer the following questions concerning two male children, a 10 and a 15 years old, each having a BMI value of 25:Table 2.19 The Mean and Standard Deviations of BMI for 10- and 15-year-old Male ChildrenBMIAgeMeanSD1016.62.31519.83.1Compute the z score for the 10-year-old.Compute the z score for the 15-year-old.Which child has a larger BMI value relative to the population of males in their age group?
62 2.62 According to the National Health Statistics Report Number 122, December 20, 2018 (Fryar, 2018), the estimated mean height of an adult male in the United States is 69 inches and the mean height of an
