Question

In: Economics

Given a population with a mean of μ= 103 and a variance of σ2 = 16,...

Given a population with a mean of μ= 103 and a variance of σ² = 16, the central limit applies when the sample size n >_25. A random sample of size n = 25 is obtained.

a. What are the mean and variance of the sampling distribution for the sample means?

b. What is the probability that X^{- >} 104 ?

c. Whta is the probability that 102 < X < 104 ?

d. What is the probability that X^- <_ 103.5 ?

Solutions

Expert Solution

It is given that the -

$\mu = 103,{\sigma ^2} = 16,{\sigma ^{}} = \sqrt {16} = 4,n = 25$ .

As the central limit theorem is applicable. Thus, Z-test is used to compute the required probability.

a) It is given that the central limit theorem is applicable. Thus, it states that the sample mean ( $\overline X$ ) and sample standard deviation ( ${\sigma _{_{\overline X }}}$ ) of the sampling distribution of means is computed as-

$\mu = \overline X = 103$ , and

${\sigma _{_{\overline X }}} = \frac{\sigma }{{\sqrt n }} = \frac{4}{{\sqrt {25} }} = \frac{4}{5} = 0.8$

b) The required probability is computed as -

$\begin{array}{c}P\left( {\overline X > 104} \right) = 1 - P\left( {\overline X < 104} \right)\\ = 1 - P\left( {\frac{{\overline X - \mu }}{{{\sigma _{\overline X }}}} < \frac{{104 - 103}}{{0.8}}} \right)\\ = 1 - \Phi \left( {1.25} \right)\\ = 1 - 0.894\\ = 0.106\end{array}$

The required value is computed using Z-table.

Therefore, the required probability is 0.106.

c) The required probability is computed as -

$\begin{array}{c}P\left( {102 < \overline X < 104} \right) = P\left( {102 < \overline X < 104} \right)\\ = P\left( {\frac{{102 - 103}}{{0.8}} < \frac{{\overline X - \mu }}{{{\sigma _{\overline X }}}} < \frac{{104 - 103}}{{0.8}}} \right)\\ = \Phi \left( {1.25} \right) - \Phi \left( { - 1.25} \right)\\ = 0.894 - 0.106\\ = 0.788\end{array}$

The required value is computed using Z-table.

Therefore, the required probability is 0.788.

d) The required probability is computed as -

$\begin{array}{c}P\left( {\overline X \le 103.5} \right) = P\left( {\frac{{\overline X - \mu }}{{{\sigma _{\overline X }}}} < \frac{{103.5 - 103}}{{0.8}}} \right)\\ = \Phi \left( {0.625} \right)\\ = 0.734\end{array}$

The required value is computed using Z-table.

Therefore, the required probability is 0.734.

Rahul Sunny answered 9 months ago

Next > < Previous

Related Solutions

Given a population with a mean of µ = 100 and a variance σ2 = 13,...

Given a population with a mean of µ = 100 and a variance σ2 = 13, assume the central limit theorem applies when the sample size is n ≥ 25. A random sample of size n = 28 is obtained. What is the probability that 98.02 < x⎯⎯ < 99.08?

Given a population with a mean of µ = 100 and a variance σ2 = 14,...

Given a population with a mean of µ = 100 and a variance σ2 = 14, assume the central limit theorem applies when the sample size is n ≥ 25. A random sample of size n = 46 is obtained. What is the probability that 98.80 < x¯ < 100.92?

Recall the probability density function of the Univariate Gaussian with mean μ and variance σ2, N(μ,σ2):...

Recall the probability density function of the Univariate Gaussian with mean μ and variance σ2, N(μ,σ2): Let X∼N(1,2), i.e., the random variable X is normally distributed with mean 1 and variance 2. What is the probability that X∈[0.5,2]?

Let X be a random variable with mean μ and variance σ2. Given two independent random...

Let X be a random variable with mean μ and variance σ2. Given two independent random samples of sizes n1 = 9 and n2 = 7, with sample means X1-bar and X2-bar, if X-bar = k X1-bar + (1 – k) X2-bar, 0 < k < 1, is an unbiased estimator for μ. If X1-bar and X2-bar are independent, find the value of k that minimizes the standard error of X-bar.

Given a variable with the following population parameters: Mean = 20. Variance = 16. a) What...

Given a variable with the following population parameters: Mean = 20. Variance = 16. a) What is the probability of obtaining a score greater than 18 and less than 21? b) What is the score at which 75% of the data falls at or below? c) What is the score at which 25% of the data falls at or above? d) Within what two scores do 95% of the scores fall (i.e symmetrically)? Bold answers.

6) Given a variable with the following population parameters: Mean = 20 Variance = 16 a)...

6) Given a variable with the following population parameters: Mean = 20 Variance = 16 a) What is the probability of obtaining a score greater than 23? b) What is the probability of obtaining a score greater than 16? c) What is the probability of obtaining a score less than 10? d) What is the probability of obtaining a score greater than 18 and less than 21? e) What is the probability of obtaining a score greater than 21 and...

Given a variable with the following population parameters: Mean = 20 Variance = 16 Sample Size...

Given a variable with the following population parameters: Mean = 20 Variance = 16 Sample Size = 35 a) What is the probability of obtaining a mean greater the 23? b) What is the probability of obtaining a mean less than 21? c) What is the probability of obtaining a mean less than 18.2? d) What is the probability of obtaining a mean greater than 19.5 and less than 21.2? e) What is the probability of obtaining a mean greater...

4) For a population has a mean of μ = 16 and a standard deviation of...

4) For a population has a mean of μ = 16 and a standard deviation of σ= 8 find the z-score corresponding to a sample mean of M= 20 for each of the following sample sizes. n=4

Use technology to construct the confidence intervals for the population variance σ2 and the population standard...

Use technology to construct the confidence intervals for the population variance σ2 and the population standard deviation σ. Assume the sample is taken from a normally distributed population. c=0.90 s=33 n=20 The confidence interval for the population variance is (_, _ ) (Round to two decimal places as needed.) The confidence interval for the population standard deviation is (_, _ ) (Round to two decimal places as needed.)

Use technology to construct the confidence intervals for the population variance σ2 and the population standard...

Use technology to construct the confidence intervals for the population variance σ2 and the population standard deviation σ2 Assume the sample is taken from a normally distributed population. c=0.980.98, s2=4.844.84, n=2727 The confidence interval for the population variance is (Round to two decimal places as needed.)

Subjects