Concept explainers
A researcher suspected that the number of betweenmeal snacks eaten by students in a day during final examinations might depend on the number of tests a student had to take on that day. The accompanying table shows joint probabilities, estimated from a survey.
Number of Snacks (Y) |
Number of Tests (X) | |||
0 | 1 | 2 | 3 | |
0 1 2 3 |
0.07 0.07 0.06 0.02 |
0.09 0.06 0.07 0.04 |
0.06 0.07 0.14 0.16 |
0.01 0.01 0.03 0.04 |
a. Find the
b. Find the probability distribution of Y and, hence, the mean number of snacks eaten by students on that day.
c. Find and interpret the conditional probability distribution of Y, given that X = 3.
d. Find the
e. Are number of snacks and number of tests independent of each other?
Trending nowThis is a popular solution!
Step by stepSolved in 4 steps
- Say someone has made a 4-sided die, and states the following probabilities of getting a side (1,2,3, or 4). In each part, state whether a proper probability distribution was used. Why or why not. a. P(1) = 0.25, P(2) = −0.10, P(3) = 0.35, P(4) = 0.25 b. P(1) = 0.25, P(2) = 0.25, P(3) = 0.25, P(4) = 0.25 c. P(1) = 0.25, P(2) = 0.10, P(3) = 0.35, P(4) = 0.25arrow_forwardIn recent years research suggests that cell phone usage during social interactions leads to distractions, which undermines the benefits of social interaction. A recent report by greater good magazine reports that a survey shows that 82% of people report that the use of cell phones during a conversation decreases the quality and what they talk about. Assuming the greater good magazine report is accurate, and then a random sample of cell phone users during A recent conversation were sampled. You are interested in the probability that a specific number of the cell phones users would report that the use of cell phones during a conversation decreases the quality of what they talk about.arrow_forwardMany track hurdlers believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on until the outermost lane, Lane 6. The data lists the number of wins for track hurdlers in the different starting positions. Calculate the chi-square test statistic x to test the claim that the probabilities of winning are the same in the different positions. Use α = 0.05. The results are based on 240 wins. Starting Position 3 4 5 6 1 2 Number of Wins 44 32 33 50 36 45 O A. 6.750 O B. 12.592 O C. 15.541 O D. 9.326 Carrow_forward
- 2.3 3 help please use diagram for answer with 5 part question (a-e) please thank you.arrow_forwardA certain company makes many, many different puzzles, all of the same difficultly level. The times it takes Jack to assemble such puzzles are normally distributed with mean 500 minutes and standard deviation 20 minutes. Sam's times are normally distributed with mean 475 minutes and standard deviation 15 minutes. Find the following probabilities.Answers to four places after the decimal, using normal charts.Each are given a copy of the same puzzle. Find the probability that Jack will finish before Sam: Find the probability that Jack's average time on 4 puzzles is less than Sam's average time on 4 puzzles:arrow_forward(first part of question is attached) (e) What is the probability that the sample proportion of students of adults living with parents or other relatives is higher than 0.3? (f) What is the probability that the sample proportion of adults living alone is between 0.1 and 0.15?arrow_forward
- Data was collected and placed into the following table. Brown eyes Blue eyes male 10 13 female 14 15 if someone who participated in the study is randomly selected, find the following probabilities. ( please use fractions and not decimals). A. P(Male B. P( female and brown eyes) C. P(Male or brown eyes) D. P( Female l blue eyes) E. P( Brown eyes l male) show work please!arrow_forwardI. Boston Children’s Hospital estimates that 90% of Americans have had chickenpoxby the time they reach adulthood.32(a) Suppose we take a random sample of 100 American adults. Is the use of the binomial distributionappropriate for calculating the probability that exactly 97 out of 100 randomly sampled Americanadults had chickenpox during childhood? Explain.(b) Calculate the probability that exactly 97 out of 100 randomly sampled American adults had chickenpoxduring childhood.(c) What is the probability that exactly 3 out of a new sample of 100 American adults have not hadchickenpox in their childhood?(d) What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox?(e) What is the probability that at most 3 out of 10 randomly sampled American adults have not hadchickenpox?arrow_forward
- A First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON