Question

In: Statistics and Probability

Consider a large population of interest. It's distribution is normal and it's mean is 187 and...

Consider a large population of interest. It's distribution is normal and it's mean is 187 and standard deviation is 112. Let X = a single observation.

(Round all probabilities to four decimals)

a) Find P(X < 192):

A random sample of 126 is taken. Let X¯ = sample average of the observations.

b) Find P( X¯ < 192):

Consider another large population of interest. It's distribution is unknown and it's mean is 187 and it standard deviation is 112. Let Y = a single observation.

A random sample of 126 is taken. Let Y¯ = sample average of the observations.

c) Find P( Y¯ < 192):

Solutions

Expert Solution

Given: The distribution of X is normal and it's mean is 187 and standard deviation is 112.

a) The required probability is:

## By using z table.

A random sample of 126 is taken. Let X¯ = sample average of the observations.

## By using z table.

Consider another large population of interest. It's distribution is unknown and it's mean is 187 and it standard deviation is 112. Let Y = a single observation.

A random sample of 126 is taken. Let Y¯ = sample average of the observations.

### Since the sample size is large the sampling distribution of sample mean is normal according to the central limit theorem.

### By using z table.

orchestra answered 1 year ago

Next > < Previous

Related Solutions

IQ scores in a large population have a normal distribution with mean=100 and standard deviation=15. What...

IQ scores in a large population have a normal distribution with mean=100 and standard deviation=15. What is the probability the sample mean for n=2 will be 121 or higher? (I understand the z score equals 1.98,could you please explain to me how the final answer is equal to 0.0239?thanks.) Also, why do we use 1 and subtract 0.9761 to get 0.0239? Where does 0.9761 come from?

Consider a normal population distribution with the value of σ known.

Consider a normal population distribution with the value of σ known.(a) What is the confidence level for the interval \(\bar{x} \pm 2.81 \sigma / \sqrt{n} ?\) (Round your answer to one decimal place.) \(\%\)(b) What is the confidence level for the interval \(\bar{x} \pm 1.43 \sigma / \sqrt{n} ?\) (Round your answer to one decimal place.) \(\%\)(c) What value of \(z_{\alpha / 2}\) in the CI formula below results in a confidence level of \(99.7 \% ?\) (Round your answer...

A population has a normal distribution with a mean of 51.4 and a standard deviation of...

A population has a normal distribution with a mean of 51.4 and a standard deviation of 8.4. Assuming n/N is less than or equal to 0.05, the probability, rounded to four decimal places, that the sample mean of a sample size of 18 elements selected from this population will be more than 51.15 is?

The distribution of systolic blood pressure in the general population is normal with a mean of...

The distribution of systolic blood pressure in the general population is normal with a mean of 130 mm Hg and a standard deviation of 20 mm Hg. In a special subgroup of 85 people with glaucoma, we find that the mean systolic blood pressure is 135 m Hg with a standard deviation of 20 mm Hg. (a) Assuming that the standard deviation of the glaucoma patients in the same as that of the general population, test for an association between...

A population has a normal distribution with a mean of 51.5 and a standard deviation of...

A population has a normal distribution with a mean of 51.5 and a standard deviation of 9.6. Assuming , the probability, rounded to four decimal places, that the sample mean of a sample of size 23 elements selected from this populations will be more than 51.15 is:

A population forms a normal distribution with a mean of µ = 120 and a standard...

A population forms a normal distribution with a mean of µ = 120 and a standard deviation of σ = 14. If two scores were selected from this population, how much distance would you expect, on average, between the second score and the population mean? A sample of n = 20 scores from this population has a mean of M = 90, do you think this sample is relative typical or extreme to the population? Explain.

Consider a distribution of student scores that is Normal with a mean of 288 and a...

Consider a distribution of student scores that is Normal with a mean of 288 and a standard deviation of 38. 1. What is the normalized value (z-score) of a score of 300? 2. What is the proportion of students with scores greater than 300? 3. What is the proportion of students with scores between 290 and 320? 4. Using the 68-95-99.7 rule, what are the two scores symmetrically placed around the mean that would include 68% of the observations?

For a population with a normal distribution with mean 23.5 and a standard deviation 2.6, determine...

For a population with a normal distribution with mean 23.5 and a standard deviation 2.6, determine the following. (a) About 68% of the population lies between what limits? (b) About 95% of the population lies between what limits? (c) About 99.7% of the population lies between what limits

A population parameter has a normal distribution and has a mean of 45 and variance of...

A population parameter has a normal distribution and has a mean of 45 and variance of 15. From this population a sample is selected with a size of 19 and the variance of the sample is 17. Does this sample support the population variance? Evaluate.

a. A random sample of 25 is taken from a normal distribution with population mean =...

a. A random sample of 25 is taken from a normal distribution with population mean = 62, and population standard deviation = 7. What is the margin of error for a 90% confidence interval? b. Repeat the last problem if the standard deviation is unknown, given that the sample standard deviation is S=5.4

Subjects