Data are provided below for the weight of dogs in pounds at a rescue center. Mak
ID: 3322644 • Letter: D
Question
Data are provided below for the weight of dogs in pounds at a rescue center. Make a histogram the distribution of the weights of the dogs in pounds. Using StatCrunch, make a histogram for to show the pro the data. Attach the graphs to your exam with the problem labeled. 51 65 67 69 70 71 73 74 74 75 75 75 77 78 78 79 79 80 80 80 81 81 82 82 83 83 84 85 85 88 18. Describe the distribution displayed in your histogram from problem 17. Use correct grammar, punctuation, and complete sentences in your descriptions. 19. Birth weights of babies in the United States can be modeled by a normal distribution with mean of 7.17 pounds and standard deviation 1.21 pounds. Those weighing less than 5.51 pounds are considered to be of low birth weight. What is the probability that a randomly selected baby in the United States will be between 6 and 8 pounds? from a standard deck of playing cards. Find the probability of choosing a 20. A card is being drawn heart or not a jack. 21. Suppose that the probability that a flight leaving from Cleveland to Chicago arrives on time 75% of the time. Assume that this flight follows a binomial distribution and that a sample of 18 flights is selected. Approximately, how many trips would you expect to arrive on time out of 18 randomly selected flights, give or take how many flights? Round to the nearest whole flight.Explanation / Answer
Q19.
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 7.17
standard Deviation ( sd )= 1.21
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 6) = (6-7.17)/1.21
= -1.17/1.21 = -0.9669
= P ( Z <-0.9669) From Standard Normal Table
= 0.1668
P(X < 8) = (8-7.17)/1.21
= 0.83/1.21 = 0.686
= P ( Z <0.686) From Standard Normal Table
= 0.7536
P(6 < X < 8) = 0.7536-0.1668 = 0.5868
Q20.
chosing a heart from a pack of cared = 13/52
chosing a card not a jack is = 1 - (4/52) = 48/52
being a heart and not a jack = 12/52
P( heart or not a jack) = 13/52 + 48/52 - 12/52 = 49/52 = 0.9423
Q21.
BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 18 * 0.75
= 13.5 ~ 14 will arrive
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.