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

7. μ=204.9 and σ=5.4968

8. μ=75.9 and σ=0.7136

9. p=0.9452

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=(\sigma)/(√(n))\\\\=(81.9)/(√(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=(\sigma)/(√(n))\\\\=(9.6)/(√(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=(\sigma)/(√(n))\\\\=(88)/(√(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>(117.4-\mu_(\bar x))/(\sigma_x))\\\\=P(Z>(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