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Tbeimal 2 Find decimal The Score: probability the = places where 0 represents recent study 5.2.7-T 0.25 Homework: that 2 : probability 2 needed:) probabiydoata that...

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Tbeimal 2 Find decimal The Score: probability the = places where 0 represents recent study 5.2.7-T 0.25 Homework: that 2 : probability 2 needed:) probabiydoata that a S randomly randomly J,906 yncorigd lynaorcigd and hspineibie life Section selected I and participants study participant's study study participant's study were of 8 01 devepiorsenz 2 anrdcpaptnse Il the 01 paxse complete) was response was less response H Homework between was was Answer parts (a) (d) below: life . their tha

Tbeimal 2 Find decimal The Score: probability the = places where 0 represents recent study 5.2.7-T 0.25 Homework: that 2 : probability 2 needed:) probabiydoata that a S randomly randomly J,906 yncorigd lynaorcigd and hspineibie life Section selected I and participants study participant's study study participant's study were of 8 01 devepiorsenz 2 anrdcpaptnse Il the 01 paxse complete) was response was less response H Homework between was was Answer parts (a) (d) below: life . their than 4 between less The current lives 1 and 6 is 60,687 responses - Score: and 6. on a 21 jl (Round to four Question Help 8 (Round to four V 1.75 Sava



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For Exercises $2-4,$ use the following information. Mr. Bash gave a quiz in his social studies class. The scores were normally distributed with a mean of 21 and a standard deviation of $2 .$
What is the probability that a student chosen at random scored between 17 and 25$?$

For this problem. We're told that 2% of two million high school students would take the S A T every year, requires special accommodations and then were asked to consider a random sample of 25 students who have recently taken the test. And we're asking probabilities so weakened we can define X as the number of successes being the number of students who required special accommodation and because we're sampling without replacement from the population. The samples are not strictly independent, but because 25 is far less than 5% of the total population, X can be pretty reliably estimated as a binomial random variable based on 25 size of 25 and the probability of success of 0.2 part A were asked, What is the probability that exactly one required special accommodation? What is the probability that we have exactly one success? So that's the probability mass function for the binomial, and this comes out to approximately zero point 308 now for B were asked, what is the probability that at least one received special accommodation? So what is the probability that we have at least one success we can rewrite this as one minus the probability of getting exactly zero successes, and this comes out to approximately zero point 397 or C. We're looking for the probability of at least two successes, so this could be rewritten as one minus probability. That X is less than or equal to one, and this equals one minus 0.911 which equals 0.89 for a party, you were asked, what is the probability that the number of successes is within two standard deviations of the mean number of successes or the expected number of successes? So to calculate this, we must calculate the mean number of successes where this distribution and the standard deviation So for a binomial random variable, the mean or expected value for the number of successes is he going to end terms? P, which comes out to you 1/2 and the standard deviation of a binomial random variable is given by the square root of N times. P Times Q, which comes out to 0.7, so within two standard deviations of the expected value. We're within this range, so the mean value is 0.5 and so plus or minus two times 0.7, which is the standard deviation, and that gives us a range from minus 0.9 to 1.9. So that means we're looking for the probability of a certain number of successes. So we for outcomes we can only have integer numbers of students, and that can't be negative. So this would be replaced by zero Can't be any lesson zero, and the number two would be more than two standard deviations greater than the expected number. So therefore, we must look for outcomes that are less than or equal to one. So we're looking for the probability that the number of successes is between zero and one, which is the same as saying we're looking for the probability of at most one success, which comes out to 0.911 and finally for party. It's explained to us that students who receive special accommodation are allowed 4.5 hours for the S A T, and students who do not are allowed three hours to rate the S A T. So we're asked, what do we expect? The average time allowed for the 25 students to be, so we can begin by writing a function that defines the average time spent by the 25 students. So for a student who is given special accommodation to get 4.5 hours, so we have 4.5 hours times number of students who get the special accommodation and then the students who don't get three hours and the number of them will be 25 minus the number of successes. And to make it an average, we must divide by the number of students, divide by 25. So this function defines the average number of hours that the students are given for the S A. T out of a sample of 25 and this can be simplified as 75 plus 1.5 x over 25. So for the question, we're asked, what is the number of hours allocated to the students on average, that we expect. So what is their expected number for the average number of hours allocated to the students? So what is asking for us? The expected value of the function h of X. So we can say that is equal to the expected value of 75 plus 1.5 x over 25. And because of the linearity of expectation, we can factory with the 1/25 and rewrite the rest of it like this. Now we have 1/25 so the expected value of a constant is just that constant. So we have 75 plus 1.5 times the expected value for X. And we've already calculated that previously, as 0.5. So this comes out to 3.3 So are we expect that the average number of hours allocated to the students to the 25 students writing S A T will be 3.3 hours.

In this exercise, the random variable is a number of questions that the students answers correctly, So n equals 15, and we know that this number follows a bell nominal distribution compared to a one wrong answer is eliminated, so there are four answers left And the probability is 1/4. We can use the formula of the phenomenal distribution to compute the distribution of the random variable, and then for about a, the answer is just the sum of the probability from P 10 to p 15, and they have computed their numbers, so the answer is approximately points 000795 for per B. two wrong answers are eliminated and has P equals 1 3rd. We can still compute the probabilities And the some of the probability from 10 to 15 is approximately 0.0 850

Hello, everyone in this video. We're going to look at frequency distributions to find probability. Here's the table. Make sure to add all these numbers up to find the total. This total will be useful for the denominator. The denominators of our probabilities. The total is 500. Which is also given in the problem. Top 500 ceos. Next we'll move on to part A. Which is a ceo of either B. S. Or B. A. So which is the second row 1 64. Over a total of 500. I should give us 0.3 to eight. You want to convert that percentage? Remember? Move the decimal two places to the right. 32.8%. Oh 32.8% of R 500. C E o s have A B. S. Or B A. Next C E o S with either a N B A. Or a J. D. Which is 1 94. I mean, sorry, 1 91 plus 50 over 500. Actually give us 0.482 or 48.2%. So 48% of our top 500 Ceos have NBA'S or JDS. Next we look at at least some college degree. That means all over Ceos not including the first one because 14 of them do not have any college experience but no college degrees. Sorry. So 1 64 plus 1 91. That's 50 that's 81 over 500. We have 0.972 Just roughly. No sorry not roughly. When you move the uh that's the most two places to the right you should get 97 2%. So 97.2% over 500 Ceos have at least um a college degree. It

For this problem. We are told that the average score for a science reasoning test is 20.8 and that the standard deviation is 4.6 and that we can assume the scores are normally distributed. So we have a mean of 20.8 and the standard deviation of 4.6. And for part a were asked what the probability is that a randomly selected student with score at least 25. That's the probability of the score in greater than or equal to 25. So if we want to standardize this, are converting to set scores through this formula. So we can say that this is equal to the probability that said, it's greater than 25 minus 20.8 over 4.6, and that is equal to one minus the probability that said is less than 0.913 and that probability comes out to 0.181 So for Part B were asked, what is the probability that randomly selected students scores between 20 and 26? So that is probability that scores between 20 and 26 and converting to the said domain that is the same as asking What is the probability? At 20 minus 20.8 over 4.6 his listens at, which is less than 26 minus 20.8 over 4.6. And this probability comes out to you 0.87 minus zero point for three, which equals 0.440 And then, in part, C finally were asked, What is the probability that a randomly select selected students scores less than 16? That is the probability that is less than 16 minus 20.8 over 4.6, which is equal to the probability that set is less than minus 1.43 and that probability is equal to 0.148


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