Scores for men on the verbal portion of the SAT-I test are normally distributed with a mean of 509 and a standard deviation of 112.
(a) If 1 man is randomly selected, find the probability that his score is at least 579.
(b) If 13 men are randomly selected, find the probability that their mean score is at least 579.
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and do all kinds of neat Z-table stuff. Just enter your mean and std and you can see the areas (probabilities) for various kinds of calculations.
To solve these two problems, we can use the concept of z-scores. The z-score represents the number of standard deviations an observation is from the mean.
(a) To find the probability that a man's score is at least 579, we need to find the area under the normal curve past the z-score of the corresponding score.
First, we need to calculate the z-score for a score of 579:
z = (x - mean) / standard deviation
z = (579 - 509) / 112
z = 0.625
Next, we can use a standard normal distribution table or a calculator to find the probability associated with this z-score. The standard normal distribution table provides the probability to the left of a given z-score. Since we want to find the probability to the right of 0.625, we can subtract the probability to the left of 0.625 from 1.
P(X ≥ 579) = 1 - P(Z < 0.625)
Referring to the standard normal distribution table, the probability associated with a z-score of 0.625 is approximately 0.7335.
P(X ≥ 579) = 1 - 0.7335 = 0.2665
Therefore, the probability that a randomly selected man's score is at least 579 is approximately 0.2665, or 26.65%.
(b) To find the probability that the mean score of 13 men is at least 579, we can use the Central Limit Theorem. According to the Central Limit Theorem, when a sample size is sufficiently large (typically n > 30), the distribution of sample means will be approximately normally distributed, even if the original population is not.
Since the population distribution is already assumed to be normal, we can use the properties of normal distribution to find the probability that the mean score is at least 579.
The mean of the sample means will still be the same as the population mean (509), and the standard deviation of the sample means, also known as the standard error, can be calculated by dividing the population standard deviation (112) by the square root of the sample size (√13).
Standard error = standard deviation / √n
Standard error = 112 / √13
Standard error ≈ 30.89
Now, we can calculate the z-score for a sample mean of 579:
z = (x - mean) / standard error
z = (579 - 509) / 30.89
z ≈ 2.27
Next, we can find the probability associated with this z-score using a standard normal distribution table or calculator.
P(X̄ ≥ 579) = P(Z ≥ 2.27)
Referring to the standard normal distribution table, the probability associated with a z-score of 2.27 is approximately 0.9893.
P(X̄ ≥ 579) = 1 - P(Z < 2.27)
P(X̄ ≥ 579) = 1 - 0.9893
P(X̄ ≥ 579) ≈ 0.0107
Therefore, the probability that the mean score of 13 men is at least 579 is approximately 0.0107, or 1.07%.
To find the probability in both cases, we need to use the z-score formula and the standard normal distribution table.
(a) To find the probability that the score of one man is at least 579, we need to calculate the z-score for this score and then find the probability associated with that z-score.
The z-score formula is: (X - μ) / σ, where X is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
In this case:
X = 579
μ = 509
σ = 112
Substituting these values into the formula, the z-score is: (579 - 509) / 112 = 0.625.
Next, we need to find the probability associated with this z-score using the standard normal distribution table. The table provides the probabilities for a range of z-scores up to a certain value.
Looking up the z-score of 0.625 in the standard normal distribution table, we find that the corresponding probability is 0.7340.
However, we are interested in the probability of the score being AT LEAST 579, so we need to find the probability of the score being below 579 and subtract it from 1.
So, the probability that the score of one man is at least 579 is: 1 - 0.7340 = 0.2660.
(b) To find the probability that the mean score of 13 men is at least 579, we use the concept of the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means from a population with any distribution approaches a normal distribution as the sample size increases.
In this case, we have a sample size of 13. We know the mean and standard deviation of the population, so we can calculate the standard deviation of the sample mean (also known as the standard error).
The formula for the standard error is: σ / sqrt(n), where σ is the population standard deviation and n is the sample size.
In this case:
σ = 112 (population standard deviation)
n = 13 (sample size)
Substituting these values into the formula, the standard error is: 112 / sqrt(13) = 31.334.
Next, we calculate the z-score for the mean score of 579 using the formula for the z-score for the sample mean: (X - μ) / (σ / sqrt(n)).
In this case:
X = 579
μ = 509
σ = 31.334
n = 13
Substituting these values into the formula, the z-score is: (579 - 509) / (31.334 / sqrt(13)) = 2.704.
Using the standard normal distribution table, the probability associated with a z-score of 2.704 is approximately 0.9966.
Again, we are interested in the probability that the mean score is AT LEAST 579, so we need to find the probability of the mean score being below 579 and subtract it from 1.
Therefore, the probability that the mean score of 13 men is at least 579 is: 1 - 0.9966 = 0.0034.