**Sampling Distribution of Proportion**Suppose a simple random sample of size \( n = 1000 \) is obtained from a population whose size is \( N = 2,000,000 \) and whose population proportion with a specified characteristic is \( p = 0.55 \). Complete parts (a) through (c) below.**(a) Describe the sampling distribution of \(\hat{p}\).**- A. Approximately normal, \(\mu_{\hat{p}} = 0.55\) and \(\sigma_{\hat{p}} \approx 0.0004\)- B. Approximately normal, \(\mu_{\hat{p}} = 0.55\) and \(\sigma_{\hat{p}} \approx 0.0157\)- C. Approximately normal, \(\mu_{\hat{p}} = 0.55\) and \(\sigma_{\hat{p}} \approx 0.0002\)**(b) What is the probability of obtaining \( x = 580 \) or more individuals with the characteristic?**\[ P(x \geq 580) = \text{[ ]} \quad (\text{Round to four decimal places as needed.}) \]**(c) What is the probability of obtaining \( x = 530 \) or fewer individuals with the characteristic?**\[ P(x \leq 530) = \text{[ ]} \quad (\text{Round to four decimal places as needed.}) \]

a) Consider that p is population proportion. It is given that, p = 0.55.Then the point estimate of p is p̂ = x/n, such that, x is the number of favorable events and n is number of all of the events. Assume that, p̂ denotes the sample proportion. The mean of the sampling distribution of the sample proportion for sample size of 1,000 is, µp̂ = p = 0.55. The sample standard deviation of the sampling distribution of the sample proportion for sample size of 1000 is, σp̂ = √[p (1 – p)/n] = √[0.55 (1 – 0.55)/1000] = 0.0157. The distribution of p̂ follows approximately normal distribution with mean 0.55 and standard deviation of 0.0157. Correct option is B.b) The standard normal variable Z is defined as, z = (p̂ – p)/ √[p (1 – p)/n] where p is the claimed population proportion, p̂ is the sample proportion, and n is the sample size. The probability of obtaining x = 580 or more individuals with the characteristic is, Thus, the probability of obtaining x = 580 or more individuals with the characteristic is 0.0281.c) The probability of obtaining x = 530 or fewer individuals with the characteristic is, Thus, the probability of obtaining x = 530 or fewer individuals with the characteristic is 0.1020.

Math

Statistics

Suppose a simple random sample of size n= 1000 is obtained from a population whose size is N = 2,000,000 and whose population proportion with a specified characteristic isp=0.55. Complete parts (a) through (c) below. (a) Describe the sampling distribution of p. O A. Approximately normal, µn = 0.55 and on z 0.0004 "p O B. Approximately normal, Ha = 0.55 and on 2 0.0157 OC. Approximately normal, µa = 0.55 and on 0.0002 (b) What is the probability of obtaining x = 580 or more individuals with the characteristic? P(x2 580) = (Round to four decimal places as needed.) (c) What is the probability of obtaining x = 530 or fewer individuals with the characteristic? P(xs 530) = |(Round to four decimal places as needed.)

Suppose a simple random sample of size n= 1000 is obtained from a population whose size is N = 2,000,000 and whose population proportion with a specified characteristic isp=0.55. Complete parts (a) through (c) below. (a) Describe the sampling distribution of p. O A. Approximately normal, µn = 0.55 and on z 0.0004 "p O B. Approximately normal, Ha = 0.55 and on 2 0.0157 OC. Approximately normal, µa = 0.55 and on 0.0002 (b) What is the probability of obtaining x = 580 or more individuals with the characteristic? P(x2 580) = (Round to four decimal places as needed.) (c) What is the probability of obtaining x = 530 or fewer individuals with the characteristic? P(xs 530) = |(Round to four decimal places as needed.)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A population of values has a normal distribution with μ=158.8μ=158.8 and σ=91.9σ=91.9. You intend to draw a random sample of size n=141n=141.Inverse Normal CalculatorFind P10, which is the score separating the bottom 10% scores from the top 90% scores.P10 (for single values) = Find P10, which is the mean separating the bottom 10% means from the top 90% means.P10 (for sample means) = Enter your answers as numbers accurate to 1 decimal place.
8,10
let x be a ramdom variable with mean= 150 and standard deviation =4.2. Ramdom samples of fixed size n=30 are drawn from the distribution of x. determine the distribution, mean, and standard deviation of sample mean
A population has a mean µ = 161 and a standard deviation o = 20. Find the mean and standard deviation of the sampling distribution of sample means with sample size n = 56. The mean is = μx (Round to three decimal places as needed.) and the standard deviation is o- = X 3 X
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…
By measuring the amount of time it takes a component of a product to move from one workstation to the next, a technologist has estimated that the standard deviation is 5.00 seconds. What sample size should be used in order to be 95% certain that the mean transfer is estimated to within ±1.00 seconds?
34) If a population exhibits a skew of -5.00 and an excess kurtosis of 5.00 which of the following is true? The population mean exceeds the median. The population variance exceeds the standard deviation. The population standard deviation exceeds the variance. The population median equals the mean. The population median exceeds the mean.
10
Question 7 > You wish to test the following claim (Ha) at a significance level of a = 0.05. H.: 41 Ha: µ1 7 42 You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you have reason to believe that the variances of the two populations are equal. You obtain a sample of size nį = 14 with a mean of M1 first population. You obtain a sample of size n2 of SD2 = 7.4 from the second population. = 67.8 and a standard deviation of SD1 19 with a mean of M2 15.5 from the 61.7 and a standard deviation In the answer box below insert 2.0395 as the critical value for this test. (Report answer accurate to three decimal places.) critical value What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = %3D The test statistic is... O in the critical region O not in the critical region