Compute the following probabilities for thestandard normal distribution Z.А. Р(0 -1.1)-

We have compute the following using standard normal table

Answered: Compute the following probabilities for…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The test scores on a placement exam are normally distributed random variables with a mean of 104 and standard deviation of 12. 12) Find the probability that a selected test score is less than 117 [[0.8599]] 13) Find the probability that a selected test score is between 92 and 122 [[0.7745]] 14) Find the placement test score that has 75% of the other scores below it. [[104 + 0.67(12) = 112.04]]
If the random variable X is normally distributed with a mean equal to .45 and a standard deviation equal to .40, then P(X ≥ .75) is: (a) 0.2266 (b) 0.4525 (c) 0.7734 (d) 0.7500 (e) None of the above
Suppose X is a random variable of uniform distribution between 1 and 7. Find E(X)
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Compute the following probabilities for the standard normal distribution Z. А. Р(0 -2)=
Let the random variable X denote the length of time (in seconds) required to complete a certain task. Suppose that the completion times are normally distributed with a mean of 216 seconds and a standard deviation of 50 seconds. (i) Find the probability that the time needed to complete the task is between 260 and 290 seconds. (ii) Find the value of k if P(X < k) = 0.025. (iii) A random sample of 25 similar tasks is selected. Find the probability that the mean completion time will be more than 200 seconds.
Fifteen items or less: The number of customers in line at a supermarket express checkout counter is a random variable with the following probability distribution. X 0 1 2 3 4 5 P(x) 0.05 0.25 0.30 0.25 0.10 0.05 Send data to Excel
A pizza company advertises that it puts 0.5 pounds of real mozzarella cheese on its medium pizzas. Suppose that the amount of cheese on a randomly selected medium pizza is normally distributed with a mean value of 0.5 pounds and a standard deviation of 0.005 pounds. (Let y = the amount of cheese on a medium pizza. Round the answers to three decimal places.) (a) What is the probability that the amount of cheese on a medium pizza is between 0.505 and 0.510 pounds?P(0.505 ≤ y ≤ 0.510) = (b) What is the probability that the amount of cheese on a medium pizza exceeds the mean value by more than 2 standard deviations?P(y > ? + 2?) = (c) What is the probability that four randomly selected medium pizzas all have at least 0.495 pounds of cheese?
Fawns between 1 and 5 months old in Mesa Verde National Park have a body weight that is approximately normally distributed 25.36 kg and standard deviation q = 4.72 kg. Let x be the weight of a fawn in kg. What is the probability that for a fawn chosen at random: with mean u - (a) x is less than 28.88 kg? (b) x is greater than 17.34 kg? (c) x lies between 29.04 and 33.6 kg? -

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Compute the following probabilities for the standard normal distribution Z. A. P(0 < Z < 2)= В. Р(-1.6 < Z < 0.75)- С. Р(Z> -1.1)-