A random sample of 36 observations is drawn from a population with a mean equal to 66 and a standard deviation equal to 12. What is the mean and the standard deviation of the sampling distribution of x̄? μ = 66 σ = 2 Describe the shape of the sampling distribution of x̄. Does this answer depend on the sample size? The shape is that of a distribution and depend on the sample size. Calculate the z-score corresponding to a value of x̄ = 63.6. Calculate the z-score corresponding to a value of x̄ = 69.2. Find P(x̄ ≥ 63.6) (to 4 decimals) Find P(x̄ < 69.2) (to 4 decimals) Find P(63.6 ≤ x̄ ≤ 69.2) (to 4 decimals) There is a 60% chance that the value of x̄ is above (to 4 decimals).

1. Mean (μ) of x' = 66, Standard Deviation (σx') = 2.

2. Shape tends towards normal with increasing sample size.

3. Z-scores: -1.2 for x' = 63.6, 1.6 for x' = 69.2.

4. Probabilities: P(x' ≥ 63.6) ≈ 0.8849, P(x' < 69.2) ≈ 0.9452, P(63.6 ≤ x' ≤ 69.2) ≈ 0.0603, P(x' > 69.2) ≈ 0.0548.

Let's break down each part of your question step by step:

1. Mean and Standard Deviation of the Sampling Distribution of x':

  The mean of the sampling distribution of the sample mean (x') is equal to the population mean (μ), which is 66 in this case.

  The standard deviation of the sampling distribution of the sample mean (x') is equal to the population standard deviation (σ) divided by the square root of the sample size (n). So:

  Standard Deviation of x' = σ / √n = 12 / √36 = 12 / 6 = 2

2. Shape of the Sampling Distribution of x':

  The shape of the sampling distribution of the sample mean (x') tends to follow a normal distribution, especially as the sample size increases. This is known as the Central Limit Theorem. The larger the sample size, the closer the sampling distribution resembles a normal distribution.

3. Z-Scores for x' = 63.6 and x' = 69.2:

  To calculate the z-scores, you can use the formula:

  Z = (X - μ) / (σ/√n)

  - For x' = 63.6:

    Z = (63.6 - 66) / (12/√36) = (-2.4) / (2) = -1.2

  - For x' = 69.2:

    Z = (69.2 - 66) / (12/√36) = (3.2) / (2) = 1.6

4. Probability Calculations:

  - P(x' ≥ 63.6): To find this probability, you can use a standard normal distribution table or calculator. P(Z ≥ -1.2) ≈ 0.8849 (rounded to 4 decimals).

  - P(x' < 69.2): Similarly, P(Z < 1.6) ≈ 0.9452 (rounded to 4 decimals).

  - P(63.6 ≤ x' ≤ 69.2): This is the difference between the two probabilities above: P(63.6 ≤ x' ≤ 69.2) ≈ 0.9452 - 0.8849 ≈ 0.0603 (rounded to 4 decimals).

5. There is a 60% chance that the value of x' is above (to 4 decimals):

  To find the probability that x' is above a certain value, you need to calculate P(x' > 69.2). You can use the complement rule:

  P(x' > 69.2) = 1 - P(x' < 69.2) ≈ 1 - 0.9452 ≈ 0.0548 (rounded to 4 decimals).

  So, there is a 5.48% chance (rounded to 4 decimals) that the value of x' is above 69.2.

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Answer:

μ = 66, σ = 2; The distribution is bell-shaped; Yes, this depends on the sample size; z = -1.2; z = 1.6; P(X ≥ 63.6) = 0.8849; P(X < 69.2) = 0.9452; P(63.6 ≤ X ≤ 69.2) = 0.8301; 65.5

Step-by-step explanation:

The central limit theorem states that if the sample size is greater than 30, the sample mean is roughly the same as the population mean.  This means it is 66.

The standard deviation of a sampling distribution of means is given by

σ/√n

For our data, this is

12/(√36) = 12/6 = 2

The central limit theorem states that the sampling distribution is approximately normal, so it will be bell-shaped.

The formula for the z score of a sampling distribution of means is

For the value of x = 63.6,

z = (63.6-66)/(12/(√36)) = -2.4/2 = -1.2

For the value of x = 69.2,

z = (69.2-66)/(12/(√36)) = 3.2/2 = 1.6

Using a z table, we see that the area under the curve to the left of z = -1.2 (for x = 63.6) is 0.1151.  However, we want P(x̄ ≥ 63.6); this means we want the area to the right.  We subtract our value from 1:

1-0.1151 = 0.8849

Using a z table, we see that the area under the curve to the left of z = 1.6 (for x = 69.2) is 0.9452.  This is P(x̄ < 69.2).

Since we have the area under the curve to the left of each endpoint, to find P(63.6 ≤ x̄ ≤ 69.2) we subtract these values:

0.9452-0.1151 = 0.8301

To find the value that would correspond in 60% of values being larger than, we first consider the fact that the z table gives us areas to the left of values, which is probabilities less than the value.  Our question is what number has a probability of 60% being larger than; this means we need to subtract from 1:

1-0.6 = 0.4

In a z table, we find the value as close to 0.4 as we can get.  This is 0.4013, which corresponds with a z score of -0.25.

Substituting this into our z formula, we have

Multiply both sides by 2:

2(-0.25) = ((X-66)/2)(2)

-0.5 = X-66

Add 66 to each side:

-0.5+66 = X-66+66

65.5 = X