mean 176 and standard deviation, σ. (Case A) Repeatedly take samples of size 28 from this population and in each case compute the sample mean, .  Let x-bar(A)and sigma(A)  be respectively the mean and   the standard deviation of these sample means. (Case B) Repeatedly take sample of size 64 from this population and in each case compute the sample mean, .  Let x-bar(B) and sigma(B)  be respectively the mean and   the standard deviation of these sample means. 1)Choose the most appropriate option.  sigma(A) greater than sigma(B) sigma (A) not less than sigma (B) sigma (A) lessthan sigma (B) sigma (A)= sigma (B) 2)If the population standard deviation, σ, is 32, then 95% of the sample means in case B fall in this interval. (168,184), (171,181),(164,188),(172,180) 3)In case B, the percent of the sample means that are less than 177.2 is 0.6179, 0.7376, 0.7257, 0.7179

a distribution with a mean of u=41 and a standard deviation of 4 is transformed into a standardized distribution with a mean of u=100 and a standardized deviation =20, find the new standardized score for each of the following values from the original population, x=39, x=36, x=45, x=50

A variable of two populations has a mean of 30 and a standard deviation of 18 for one of the populations and a mean of 30 and a standard deviation of 32 for the other population. Complete parts (a) through (c). a. For independent samples of size 9 and 16, respectively, find the mean and standard deviation of X₁ -X₂. (Assume that the sampling is done with replacement or that the population is large enough.) The mean of X₁ -X₂ is. (Type an integer or a decimal. Do not round.) The standard deviation of X₁-X₂ is (Round to four decimal places as needed.) b. Must the variable under consideration be normally distributed on each of the two populations for you to answer part (a)? Choose the correct answer below. OA. No, the variable does not need to be normally distributed for the formulas for the mean and standard deviation of x₁-x₂ to hold as long as the sample sizes are large enough, as long as the sampling is done with replacement. B. No, the formulas for the mean and standard deviation of…

i was wondering if i could get help with questions a and b

Some plastic rods are automatically cut into nominal lengths of 6 inches.  Actual lengths are normally distributed about a mean of 6 inches and their standard deviation is 0.06 inch. What proportion of the rods exceeds tolerance limits of 5.9 inches and 6.1 inches? To what value does the standard deviation need to be reduced if 99% of the rods must be within tolerance?

Let a population consist of the values 11 cigarettes, 12 cigarettes, and 22 cigarettes smoked in a day. Show that when samples of size 2 are randomly selected with replacement, the samples have mean absolute deviations that do not center about the value of the mean absolute deviation of the population. What does this indicate about a sample mean absolute deviation being used as an estimator of the mean absolute deviation of a population? Calculate the mean absolute deviation for each possible sample of size 2 from the population. Mean Absolute Deviation Sample {11,11} {11,12} {11,22} {12,11} {12,12} {12,22} {22,11} {22,12} {22,22} (Type integers or decimals rounded to one decimal place as needed.) Budget Pi nd Pow....P Enter your answer in the edit fields and then click Check Answer. Clear All Check Answer parts

Suppose that the duration of a particular type of criminal trial is known to be normally distributed with a mean of 20 days and a standard deviation of 5 days. Let X be the number of days for a randomly selected trial. Round all answers to 4 decimal places where possible.