Use Excel/Megastat to create the density probability plot of normal distribution. Take μ = 50and σ = 3.5. Set up the range values of x from 40 to 80.
Hint: Use norm.dist function and sketch the curve with tick = 0.1
Use Excel/Megastat to create the density probability plot of normal distribution. Take μ = 50and σ...
For a normal distribution, find the probability of being (a) Between μ−3σ μ − 3 σ and μ+3σ μ + 3 σ (b) Between 2 standard deviations below the mean and 2.5 standard deviations above the mean (c) Less than μ−1σ μ − 1 σ Use the Standard Normal Table in your textbook or Excel to obtain more accuracy.
Given a normal distribution with μ=55 and σ=3.0, a) What is the probability that X greater than>51? B) What is the probability that X less than<49? c) For this distribution,99%of the values are less than what X-value? d) Between what two X-values (symmetrically distributed around the mean) are 80% of the values?
. IQ scores have a normal distribution with μ = 90 and σ = 10. a) Find the probability for a score over 100. b) Find the score needed for the top 5%. c) Find the probability that the mean of 10 scores is under 80. (Hint: use CLT)
Let fy(x, μ, σ) stand for the probability distribution function (PDF) for the normal distribution with parameters μ and σ. Let X be a random variable with a PDF defined as follows: where t is a fixed constant between O and 1. What is E[XI? None of these
Let X have a normal distribution with mean μ and variance σ ^2 . The highest value of the pdf is equal to 0.1 and when the value of X is equal to 10, the pdf is equal to 0.05. What are the values of μ and σ?
The normal distribution has parameters μ and σ. The possible values for μ are all numbers on the real line, while σ must be positive. Write an R function that takes in a vector with two elements theta(whose values are unrestricted) and returns a list with two elements. The first element is to represents μ,mu=theta[1], and the second element is to represent σ,sigma=exp(theta[2]). Write a function that takes in a named list for mu and sigma and returns theta.
A population of values has a normal distribution with μ = 173 and σ = 36.1 Find the probability that a single randomly selected value is between 167.4 and 176.9.
A normal distribution has μ = 55 and σ = 5. What is the probability of randomly selecting a score greater than or equal to 55 from this distribution? a. About 16% b. About 84% c. About 20% d. About 22%
Given a normal distribution with (mean) μ= 50 and (standard deviation) σ = 5, what is the probability that: a) X>60 b) X<40 c) X<45 or X>65 d) Between what two values (symmetrically distributed around the mean) are ninety percent of the values?
The location of a Normal distribution is determined by its mean μ, where as its shape is determined by the standard deviation σ. To see the effect of changing μ, you are going to graph two Normal probability density functions, one with μ = 100 and another with μ = 105, both having σ = 10. Recall that for each distribution the first value should be 3σ = 30 below the mean, and the last value should be 3σ above...