Scores on a visual perception test are normally distributed with a mean of 2020 and a standard deviation of 250.
a)If one subject is randomly selected and tested, find the probability of a score greater than 1800
b)if 50 subjects are randomly selected and tested fid the probability that the sample mean is between 2000 and 2100
a) To find the probability of a score greater than 1800, we need to calculate the z-score and then find the corresponding probability.
The formula for calculating the z-score is:
z = (x - mean) / standard deviation
In this case, x = 1800, mean = 2020, and standard deviation = 250.
Substituting the values into the formula, we have:
z = (1800 - 2020) / 250
z = -8 / 250
z = -0.32
Now, we find the probability using a standard normal distribution table or calculator. The probability of a score greater than 1800 is equal to the area under the normal curve to the right of the z-score.
Using a standard normal distribution table, the probability corresponding to a z-score of -0.32 is approximately 0.6247.
Therefore, the probability of a score greater than 1800 on the visual perception test is approximately 0.6247, or 62.47%.
b) To find the probability that the sample mean is between 2000 and 2100 when 50 subjects are randomly selected and tested, we need to use the central limit theorem and the standard error of the mean.
The formula for the standard error of the mean is:
SE = standard deviation / √n
In this case, the standard deviation is 250 and the number of subjects (n) is 50. Substituting these values into the formula, we have:
SE = 250 / √50
SE = 250 / 7.0711
SE ≈ 35.3553
Next, we convert the sample mean values (2000 and 2100) into z-scores. Using the formula:
z = (x - mean) / SE
For 2000:
z1 = (2000 - 2020) / 35.3553
z1 = -0.5657
For 2100:
z2 = (2100 - 2020) / 35.3553
z2 = 2.2541
Now, we find the probabilities corresponding to these z-scores using a standard normal distribution table or calculator. The probability of a z-score of -0.5657 is approximately 0.2824, and the probability of a z-score of 2.2541 is approximately 0.9878.
To find the probability that the sample mean is between 2000 and 2100, we subtract the probability corresponding to the lower z-score from the probability corresponding to the higher z-score:
P(2000 ≤ x ≤ 2100) = P(z1 ≤ z ≤ z2) = 0.9878 - 0.2824
P(2000 ≤ x ≤ 2100) ≈ 0.7054
Therefore, the probability that the sample mean is between 2000 and 2100 when 50 subjects are randomly selected and tested is approximately 0.7054, or 70.54%.
To find the probability of a score greater than 1800 from a normal distribution, we can use the z-score formula. The z-score measures how many standard deviations an individual score is from the mean.
a) Probability of a score greater than 1800:
First, we need to calculate the z-score for a score of 1800 using the formula:
z = (x - μ) / σ
Where:
x = the value we want to find the probability for (1800)
μ = the mean (2020)
σ = the standard deviation (250)
z = (1800 - 2020) / 250
z = -220 / 250
z = -0.88
Next, we use a z-table or a statistical calculator to find the probability corresponding to the z-score. The z-table provides the area under the normal distribution curve to the left of a given z-score.
Looking up the z-score of -0.88, we find that the cumulative probability (area to the left) is 0.1884. However, we want the probability of a score greater than 1800, so we subtract the cumulative probability from 1 to get:
Probability = 1 - 0.1884
Probability = 0.8116 or 81.16%
Therefore, the probability of randomly selecting a subject who scores greater than 1800 on the visual perception test is approximately 81.16%.
b) Probability that the sample mean is between 2000 and 2100:
To find the probability that the sample mean falls within a range, we need to consider the distribution of sample means.
The distribution of sample means follows the Central Limit Theorem, which states that with a large enough sample size, the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution.
Given that we have a large enough sample size of 50, we can assume that the distribution of sample means will be approximately normal.
To find the probability that the sample mean is between 2000 and 2100, we need to convert this problem into the standard normal distribution. We will use the sample mean and the standard error of the mean.
The standard error of the mean (SEM) is calculated using the formula:
SEM = σ / sqrt(n)
Where:
σ = the population standard deviation (250)
n = the sample size (50)
SEM = 250 / sqrt(50)
SEM ≈ 35.355
We can now find the z-scores for the sample mean range using the formula:
z = (x - μ) / SEM
For 2000:
z1 = (2000 - 2020) / 35.355
z1 ≈ -0.565
For 2100:
z2 = (2100 - 2020) / 35.355
z2 ≈ 2.253
Using the z-table or a statistical calculator, we find the cumulative probabilities for z1 and z2:
P(z < -0.565) ≈ 0.2862
P(z < 2.253) ≈ 0.9871
To find the probability that the sample mean is between 2000 and 2100, we subtract the cumulative probabilities:
Probability = P(z1 < z < z2)
Probability = P(z < z2) - P(z < z1)
Probability ≈ 0.9871 - 0.2862
Probability ≈ 0.7009 or 70.09%
Therefore, the probability that the sample mean falls between 2000 and 2100 is approximately 70.09%.