A First Course in Probability (10th Edition)
10th Edition
ISBN: 9780134753119
Author: Sheldon Ross
Publisher: PEARSON
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- Researchers have created every possible "knockout" line in yeast. Each line has exactly one gene deleted and all the other genes present (Steinmetz et al. 2002). The growth rate—how fast the number of cells increases per hour—of each of these yeast lines has also been measured, expressed as a multiple of the growth rate of the wild type that has all the genes present. In other words, a growth rate greater than 11 means that a given knockout line grows faster than the wild type, whereas a growth rate less than 11 means it grows more slowly. The growth rate of a random sample of knockout lines is 0.86, 1.02, 1.02, 1.01, 1.02, 1, 0.99, 1.01, 0.91, 0.83, 1.01 What is the standard deviation and variance of growth rate for this sample?arrow_forward8. Suppose that a bacteria population doubles every eleven hours. If the initial population is 2,000 individuals, approximately how many hours would it take the population to reach 20,000?arrow_forwardSuppose that a commercial bakery produces chocolate chip cookies. Based on extensive testing, the bakery has determined that the ideal number of chocolate chips per cookie is 18. In the past, master bakers mixed the dough and spooned it onto sheet pans by hand, but the owners have decided to test a new machine. The machine extrudes dough onto baking sheets at a uniform rate. It runs continuously from 4:30 a.m. until 12:30 p.m. The number of chips per cookie depends on the viscosity of the dough, and because the dough stiffens as it sits in the machine, the machine produces cookies that contain the optimal number of chips, 18 per cookie, between the hours of 5:00 a.m. and 11 a.m. The distribution of cookies is shown in the graph below. The shaded area represents the proportion of all cookies that contain exactly 18 chocolate chips. 4:00 5:00 6:00 7:00 8:00 9:00 10:00 11:00 12:00 1:00 Compute the proportion of cookies produced each day that contain the optimal number of chocolate chips.…arrow_forward
- _) Jane bakes sourdough loaves, which are full of gluten. Unfortunately, John is allergic to gluten and wants to bake gluten-free loaves. John does not have their own “starter” (a mix of yeast, flour, and water that makes a loaf rise in the oven). Jane's starter is 50g of water and 50g of flour with gluten, and a microscopic amount of yeast that we will ignore. Jane gives John 50g of her starter. For John's first loaf, they replace the missing half of their starter with 25 g of gluten-free flour and 25g of water. This means their starter is now 25g of gluten-free flour, 25g of flour with gluten, and 50g of water. Once the yeast in this starter is fully activated, John is ready to make their first loaf. To bake the loaf they mix 50 g (half) of the starter with 225g of gluten-free flour and 225g of water, meaning the loaf is alwaysarrow_forwardFrogs have been breeding like flies at the Enormous State University (ESU) campus! Each year, the pledge class of the Epsilon Delta fraternity is instructed to tag all the frogs residing on the ESU campus. Two years ago (t = 0) they managed to tag all 42,000 of them (with little Epsilon Delta Fraternity tags). This year's pledge class discovered that all the tags had fallen off, and they wound up tagging a total of 63,000 frogs. (a) Find an exponential model for the frog population. f(t) = (b) Assuming exponential population growth, and that all this year's tags have fallen off, how many tags should Epsilon Delta order for next year's pledge class? (Round your answer to the nearest integer.) tagsarrow_forwardProbability Questionarrow_forward
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- A First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON
A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
Publisher:PEARSON