Question

A population of values has a normal distribution with μ=204.9μ=204.9 and σ=81.9σ=81.9. You intend to draw a random sample of size n=222n=222. What is the mean of the distribution of sample means? μ¯x=μx¯= (Enter your answer as a number accurate to 4 decimal places.) What is the standard deviation of the distribution of sample means? (Report answer accurate to 4 decimal places.) σ¯x=σx¯=
8)Let XX represent the full height of a certain species of tree. Assume that XX has a normal probability distribution with μ=75.9μ=75.9 ft and σ=9.6σ=9.6 ft. You intend to measure a random sample of n=181n=181 trees. What is the mean of the distribution of sample means? μ¯x=μx¯= What is the standard deviation of the distribution of sample means (i.e., the standard error in estimating the mean)? (Report answer accurate to 4 decimal places.) σ¯x=σx¯=

9) A population of values has a normal distribution with μ=135.7μ=135.7 and σ=88σ=88. You intend to draw a random sample of size n=59n=59. Find the probability that a single randomly selected value is greater than 117.4. P(X > 117.4) = Find the probability that a sample of size n=59n=59 is randomly selected with a mean greater than 117.4. P(¯xx¯ > 117.4) = Enter your answers as numbers accurate to 4 decimal places.

Answer 1

Answer:

7. μ=204.9 and σ=5.4968

8. μ=75.9 and σ=0.7136

9. p=0.9452

Step-by-step explanation:

7. - Given that the population mean =204.9 and the standard deviation is 81.90 and the sample size n=222.

-The sample mean,$\mu_x$is calculated as:

$\mu_x=\mu=204.9, \mu_x=sample \ mean$

-The standard deviation,$\sigma_x$ is calculated as:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{81.9}{\sqrt{222}}\\\\=5.4968$

8. For a random variable X.

-Given a X's population mean is 75.9, standard deviation is 9.6 and a sample size of 181

-#The sample mean,$\mu_x$ is calculated as:

$\mu_x=\mu\\\\=75.9$

#The sample standard deviation is calculated as follows:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{9.6}{\sqrt{181}}\\\\=0.7136$

9. Given the population mean, μ=135.7 and σ=88 and n=59

#We calculate the sample mean;

$\mu_x=\mu=135.7$

#Sample standard deviation:

$\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{88}{\sqrt{59}}\\\\=11.4566$

#The sample size, n=59 is at least 30, so we apply Central Limit Theorem:

$P(\bar X>117.4)=P(Z>\frac{117.4-\mu_{\bar x}}{\sigma_x})\\\\=P(Z>\frac{117.4-135.7}{11.4566})\\\\=P(Z>-1.5973)\\\\=1-0.05480 \\\\=0.9452$

Hence, the probability of a random sample's mean being greater than 117.4 is 0.9452