Suppose a simple random sample of size n= 1000 is obtained from a population whose size is N = 1,500,000 and whose population proportion with a specifiedcharacteristic isp=0.31. Complete parts (a) through (c) below.(a) Describe the sampling distribution of p.O A. Approximately normal, µa = 0.31 and on 0.0002O B. Approximately normal, Ha = 0.31 and on 0.0146O C. Approximately normal, µa = 0.31 and on 0.0004(b) What is the probability of obtaining x = 330 or more individuals with the characteristic?P(x2 330) =(Round to four decimal places as needed.)(c) What is the probability of obtaining x= 280 or fewer individuals with the characteristic?P(xs 280) = (Round to four decimal places as needed.)

Answered: Image /qna-images/answer/9f0e6265-30b2-4a89-a74d-e24db8fbaf33.jpg

Answered: Suppose a simple random sample of size…

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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let x be a normal ramdom variable with a mean =169.5 and standard deviation =4.2 ramdom samples of fixed size n=30 are drawn from the distribution, mean, standard deviation od sample mean
Suppose the lengths of the pregnancies of a certain animal are approximately normally distributed with mean u= 260 days and standard deviation o = 22 days. Complete parts (a) through (f) below. (b) Suppose a random sample of 21 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies. The sampling distribution of x is normal with H; = 260 and o: = 4.8008. (Round to four decimal places as needed.) (c) What is the probability that a random sample of 21 pregnancies has a mean gestation period of 253 days or less? The probability that the mean of a random sample of 21 pregnancies is less than 253 days is approximately 0.0724. (Round to four decimal places as needed.) Interpret this probability. Select the correct choice below and fill in the answer box within your choice. (Round to the nearest integer as needed.) O A. If 100 independent random samples of size n= 21 pregnancies were obtained from this population, we would expect sample(s) to have…
Solve for C)
A sample of size 8 was collected from an unknown population 0 1126 10 11 20 1. Use the table of expected z-scores (rounded to the nearest tenth) below to construct the normality plot with expected z-scores on the x-axis and the observations on the y-axis: 5 -1.2 -0.5 0 0.5 1.2 Clear All Draw: 22- 2 not normal normal 19 # 6 -1.3 -0.6 -0.2 0.2 0.6 1.3 7 -1.4 -0.8 -0.4 0 0.4 0.8 1.4 8 -1.4 -0.9 -0.5 -0.2 0.2 0.5 0.9 1.4 9 -1.5 -0.9 -0.6 -0.3 0 0.3 0.6 0.9 1.5 2. Based on whether the pattern above is linear or not, in your opinion, was the sample drawn from a normal population or not?
The distribution of the number of children in a household is strongly skewed right. Most households have 0, 1, or 2 children, although some have 3, 4, 5, or even more than 5 children. The mean number of children per household is μ = 2.75 children with standard deviation o = 1.22 children. Describe the shape of the sampling distribution of x for SRSS of size n = 50 from the population of households. Because n = 50 > 30, the sampling distribution of x will be approximately normal. Because n = 50 > 30, the sampling distribution of x will be skewed to the right, but not as strongly as the population. 50 > 30, the sampling distribution of x will be skewed to the right just as strongly as the population. Because n = Because n = 50 > 30, we do not have enough information to draw a conclusion about the shape of the sampling distribution of x. Because n = 50 > 30, the sampling distribution of x will be skewed to the right even more strongly than the population.
Calculate the mean and the standard deviation of the sampling distribution of possible sample proportions for each combination of sample size (n) and population proportion (p). (a) n = 336, p = 0.7 mean standard deviation (b) n = 1,344, p = 0.7 mean standard deviation
Suppose I want to sample the household incomes for EBR parish. Let's say I start with a simple random sample of 100 residents and calculate the sample mean. How would the sampling distribution of change if I increase the sample size to 1,000 residents?
5. The distribution of heights of a certain breed of terrier has a mean of 72 centimeters and a standard deviation of 10 centimeters, whereas the distribution of heights of a certain breed of poodle has a mcan of 28 centimeters with a standard deviation of 5 centimcters. Assume the distributions of both types of breeds are normally distributed. (n) Suppose we are taking a random sample of 64 terriers. What is the sampling distri- bution of mean heights of the breed of terriers? (b) Find the probability that the sample mean for a random sample of heights of 61 terriers excoedls 75 centimetcrs. (c) Find the probability that the sample mean for a random sample of heights of 64 terriers excoeds the sample mean for a random sample of heights of 100 poodles by at lenst 44.2 centimeters.
7. Ecological Applications study on the development following wildfires in the Pacific Northwest. One variable of interest to the researcher was tree diameter at breast height 110 years after the fire. The population of Douglas fir trees was shown to have an approximately normal distribution with mean tree diameter ? = 50 centimeters (cm) and ? = 14 cm. Find the diameter, d, such that 20.9% of the Douglas fir trees in the population have diameters that exceed d. published a of forests

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