### Educational Content: Discrete Uniform Population**Problem Statement:**Given the discrete uniform population characterized by the probability function $ f(x) $, find the probability that a random sample of size 96, selected with replacement, will yield a sample mean greater than 12.5 but less than 13.4. Assume the means are measured to the nearest tenth.**Probability Function:**The probability function $ f(x) $ is defined as follows:- $ f(x) = \frac{1}{3} $ for $ x = 4, 12, 20 $- $ f(x) = 0 $ elsewhere**Explanation:**This problem involves determining the likelihood of a specific range of sample means from a discrete uniform distribution. The function $ f(x) $ indicates that the population consists of three values (4, 12, and 20), each with an equal probability of $ \frac{1}{3} $. The task is to calculate the probability that the mean of a sample of 96 randomly chosen values falls between 12.5 and 13.4. **Steps Approach:**1. **Calculate the Population Mean ($ \mu $):** \[ \mu = \frac{1}{3}(4) + \frac{1}{3}(12) + \frac{1}{3}(20) = 12 \]2. **Calculate the Population Variance ($ \sigma^2 $):** \[ \sigma^2 = \frac{1}{3}((4-12)^2 + (12-12)^2 + (20-12)^2) = \frac{1}{3}(64 + 0 + 64) = \frac{128}{3} \]3. **Calculate the Standard Deviation ($ \sigma $):** \[ \sigma = \sqrt{\frac{128}{3}} \]4. **Standard Error (SE):** Since the sample size is 96, the standard error is $ \frac{\sigma}{\sqrt{96}} $.5. **Apply Z-scores to Find the Probability:** Use the Z distribution to determine the probability of the sample mean falling within the specified range.This structured approach allows for comprehensively analyzing the probability under the given discrete uniform setup.

Given The discrete uniform population to the right f(x) = 1/3 x=4,12,20 0 elsewhere also given the sample size n= 96 Find the probability of a random variable that yields the sample mean greater than 12.5 but less than 13.4Let X be a discrete random variable drawn from uniform with the probability distribution f Mean: To determine the mean use the formula of expectation E(x) = ∑xf(x) μx = 4.f(4)+12.f(12)+20.f(20) = 4.13+12.13+20.13 = 363 = 12 The mean is 12 Variance: σx2= ∑f(x)(x-μx)2 = 13((4-12)2+(12-12)2+(20-12)2) = 128/3 The standard deviation is given by σx=1283 from the obtained values of mean and SD can proceed for the sample mean The given sample size is 96 which is a large sample From the CLT theorem as the sample increases the distribution is normal Z = X-μσ/n = X-μxσx/n =X-12128/396 = X-1223 The probability of the sample mean lies between 12.5 and 13.4 is given by P(12.5<X<13.4) = =P12.5-122/3<Z<13.4-122/3 = P0.50.667<Z<1.40.667 = P(0.75<Z<2.10) = P(Z<2.10)-P(Z<0.75) = 0.9871-0.7734 = 0.2137

Math

Statistics

Given the discrete uniform population shown to the right, find the probability that a random sample of size 96, selected with replacement, will yield a sample mean greater than 12.5 but less than 13.4. Assume the means x= 4, 12, 20 f(x) = 0, elsewhere are measured to the nearest tenth.

Given the discrete uniform population shown to the right, find the probability that a random sample of size 96, selected with replacement, will yield a sample mean greater than 12.5 but less than 13.4. Assume the means x= 4, 12, 20 f(x) = 0, elsewhere are measured to the nearest tenth.

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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Problem 1P

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Concept explainers

Continuous Probability Distributions

Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.

Normal Distribution

Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!

Question

### Educational Content: Discrete Uniform Population

**Problem Statement:**
Given the discrete uniform population characterized by the probability function $ f(x) $, find the probability that a random sample of size 96, selected with replacement, will yield a sample mean greater than 12.5 but less than 13.4. Assume the means are measured to the nearest tenth.

**Probability Function:**
The probability function $ f(x) $ is defined as follows:

- $ f(x) = \frac{1}{3} $ for $ x = 4, 12, 20 $
- $ f(x) = 0 $ elsewhere

**Explanation:**

This problem involves determining the likelihood of a specific range of sample means from a discrete uniform distribution. The function $ f(x) $ indicates that the population consists of three values (4, 12, and 20), each with an equal probability of $ \frac{1}{3} $. The task is to calculate the probability that the mean of a sample of 96 randomly chosen values falls between 12.5 and 13.4.

**Steps Approach:**

1. **Calculate the Population Mean ($ \mu $):**
\[
\mu = \frac{1}{3}(4) + \frac{1}{3}(12) + \frac{1}{3}(20) = 12
\]

2. **Calculate the Population Variance ($ \sigma^2 $):**
\[
\sigma^2 = \frac{1}{3}((4-12)^2 + (12-12)^2 + (20-12)^2) = \frac{1}{3}(64 + 0 + 64) = \frac{128}{3}
\]

3. **Calculate the Standard Deviation ($ \sigma $):**
\[
\sigma = \sqrt{\frac{128}{3}}
\]

4. **Standard Error (SE):**
Since the sample size is 96, the standard error is $ \frac{\sigma}{\sqrt{96}} $.

5. **Apply Z-scores to Find the Probability:**
Use the Z distribution to determine the probability of the sample mean falling within the specified range.

This structured approach allows for comprehensively analyzing the probability under the given discrete uniform setup.

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