Scores on a national writing exam are approximately normally distributed, with a mean of 490 and a standard deviation of 20. Consider random samples of size 25 from the population of test takers.
describe the sampling distribution of a sample mean.
shape
mean=
standard deviation=
find the probability that a random sample of 25 test takers will have a mean exam score that is greater than 496
find k such that p(x<K)=.01
The sampling distribution of a sample mean is approximately normal, regardless of the shape of the population distribution, as long as the sample size is sufficiently large (typically greater than 30) and the observations are independent.
Shape: The sampling distribution of a sample mean will have a bell-shaped, approximately normal distribution.
Mean: The mean of the sampling distribution of a sample mean will be equal to the population mean. In this case, the mean of the sample means will also be 490.
Standard Deviation: The standard deviation of the sampling distribution of a sample mean, also known as the standard error, can be calculated using the formula: standard deviation / √sample size. In this case, the standard deviation of the sampling distribution will be 20 / √25, which simplifies to 4.
To find the probability that a random sample of 25 test takers will have a mean exam score greater than 496, we can use the Z-score formula and the standard normal distribution.
First, we calculate the Z-score of 496, using the formula: Z = (X - μ) / σ, where X is the sample mean, μ is the population mean, and σ is the standard deviation.
Z = (496 - 490) / 4 = 6 / 4 = 1.5
Next, we find the probability associated with a Z-score of 1.5 using a standard normal distribution table or a statistical calculator. For a Z-score of 1.5, the probability is approximately 0.9332.
Therefore, the probability that a random sample of 25 test takers will have a mean exam score greater than 496 is approximately 0.9332.
To find k such that P(X < k) = 0.01, where X is the sample mean, we can also use the Z-score formula and the standard normal distribution.
Since we're looking for the value of X that corresponds to an area of 0.01 to the left of it (P(X < k) = 0.01), we need to find the Z-score that corresponds to an area of 0.01 to the right of it (P(Z > z) = 0.01).
Using a standard normal distribution table or a statistical calculator, we find the Z-score corresponding to an area of 0.01 to the right is approximately -2.33.
Next, we can use the Z-score formula to solve for k:
Z = (X - μ) / σ
-2.33 = (k - 490) / 4
Solving for k:
-2.33 * 4 = k - 490
-9.32 = k - 490
k = -9.32 + 490
k ≈ 480.68
Therefore, k is approximately 480.68 such that P(X < k) = 0.01.
The sampling distribution of a sample mean is approximately normal, regardless of the shape of the population distribution. This is known as the Central Limit Theorem.
The mean of the sampling distribution of a sample mean is equal to the population mean. In this case, the mean is 490.
The standard deviation of the sampling distribution of a sample mean, also known as the standard error, is equal to the population standard deviation divided by the square root of the sample size. In this case, the standard deviation is 20 divided by the square root of 25, which is 4.
To find the probability that a random sample of 25 test takers will have a mean exam score greater than 496, we need to convert this to a z-score using the formula:
z = (x - μ) / (σ / √n)
where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x = 496, μ = 490, σ = 20, and n = 25.
z = (496 - 490) / (20 / √25)
z = 6 / (20 / 5)
z = 6 / 4
z = 1.5
To find the probability of getting a z-score greater than 1.5, we can use a standard normal distribution table or a calculator to find the corresponding probability. The probability is approximately 0.0668, or 6.68%.
To find the value of k such that the probability of getting a score less than k is 0.01, we can use a standard normal distribution table or a calculator. The corresponding z-score is approximately -2.33.
Using the formula for z-score:
z = (x - μ) / (σ / √n)
we can rearrange it to solve for x:
x = z * (σ / √n) + μ
In this case, z = -2.33, μ = 490, σ = 20, and n = 25.
x = -2.33 * (20 / √25) + 490
x = -2.33 * 4 + 490
x = -9.32 + 490
x = 480.68
Therefore, k is approximately 480.68.