A random variable represents the number of successes in 20 Bernoulli trials, each with probability of success p = 0.55.(A) Find the mean and standard deviation of the random variable.(B) Find the probability that the number of successes lies within 1 standard deviation of the mean.(A) The mean of the random variable is 11. (Type an integer or a decimal.)The standard deviation of the random variable is approximately(Round to three decimal places as needed.)

The number of Bernoulli trials is .The probability of success is .There is a random variable , that…

Answered: Find the mean and standard deviation of…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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The IQ's in a certain population are normally distributed with a mean score of 104 and a standard deviation of 14. If we select a person at random from this population, find the probability that their IQ will be: 1. Find the 67th percentile of IQ's in this population. 2. Find the 20 The percentile of IQ's in this population 3. Find the IQR of IQ's in this population.
Susan jogs to the park each day and the trip takes an average of µ = 22 minutes. The distribution of jog times is approximately normal with a standard deviation of σ = 10 minutes. For a randomly selected day, what is the probability that Susan’s jog to the park will take less than 16 minutes? Select one: a. 0.2743 b. 0.8957 c. 0.3236 d. 0.7554
For a certain group of individuals, the average heart rate is 74 beats per minute. Assume the variable is normally distributed and the standard deviation is 4 beats per minute. If a subject is selected at random, find the probability that the person has the following heart rate. Use a graphing calculator. Round the answers to four decimal places. Less than 78 beats per minute =
Cody took his first physics exam and scored an 80. The population mean for this exam is 70, and the standard deviation is 5. What is the probability of selecting a person with a score greater than Cody’s?
A surgery has a success rate of 90%. Suppose that the surgery is performed on threepatients Graph the probability distribution for X using a histogram.Find the mean of X.Find the variance and standard deviation of X.
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 206.8 cm and a standard deviation of 0.5 cm. Find the probability that the length of a randomly selected steel rod is greater than 208.4 cm. P(x > 208.4 cm) = P(z > Shade: Left of a value Click and drag the arrows to adjust the values. -4 -3 -2 -1 2. -1.5
Assume time to failure density function (in months) of transplanted kidneys has a Weibull distribution with shape factor of 3 and scale of 6. Use R and answer the following questions (enter your answers with 3 decimal points):
Assume that adults have IQ scores that are normally distributed with a mean of u = 105 and a standard deviation o = 15. Find the probability that a randomly selected adult has an IQ between 89 and 121. Click to view page 1 of the table. Click to view page 2 of the table. Standard Normal Table (Page 2) The probability that a randomly selected adult has an IQ between 89 and 121 is (Type an integer or decimal rounded to four decimal places as needed.) POSITIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 5279 .5319 .5359 0.1 .5398 .5438 5478 .5517 .5557 .5596 ,5636 ,5675 .5714 .5753 0.2 .5793 .5832 5871 5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 6950 .6985 7019 .7054 .7088 7123 7157 .7190 .7224 0.6 .7257 .7291 7324 7357 7389 7422 7454 7486…
The distances golf balls travel under laboratory testing conditions are normally distributed. The mean is 312 yards. The probability the golf ball travels further than 319 yards is 0.28. Find the probability a randomly selected golf ball travels between 305 yards and 319 yards. Express the answer as a decimal value rounded to the nearest hundredth.

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