Question

7)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

Q.7) Given that, a population of values has a normal distribution with μ=204.9 and σ=81.9

Sample size ( n )= 222

The mean and standard deviation of the sampling distrbution of sample mean are,

Q.8) Given that, a population of values has a normal distribution with μ = 75.9 and σ = 9.6

sample size ( n ) = 181

The mean and standard deviation of the sampling distrbution of sample mean are,

Q.9) Given that, a population of values has a normal distribution with μ = 135.7 and σ = 88

We want tp find, P(X > 117.4)

Therefore, probability is 0.5832

sample size ( n ) = 59

The mean and standard deviation of the sampling distrbution of sample mean are,

We want to find,

Therefore, the probability is 0.9452