SAT scores are normally distributed. The SAT in English has a mean score of 500 and a standard deviation of 100.
a. Find the probability that a randomly selected student's score on the English part of the SAT is between 400 and 675.
b. What is the minimum SAT score that a student can receive in order to score in the top 10%?
c. Forty-nine students are sampled. Find the probability that their mean score for the SAT is less than 470.
I have the following answers, but do not know how they were derived:
a. 0.8013
b. 628
c. 0.0179
Here's a few hints to get you started:
a. Find the z-scores. Use the formula: z = (x - mean)/sd
z = (400 - 500)/100 = ?
z = (675 - 500)/100 = ?
Once you have the two z-scores, look at the z-table to determine the probability between those two scores.
b. Check the z-table to determine the top 10%. Use that value for z. Use the z-score formula. You will have z, the mean, and the standard deviation. Solve the formula for x.
c. Formula: z = (x - mean)/(sd/√n)
With your data:
z = (470 - 500)/(100/√49) = ?
Once you have the z-score, check the z-table for the probability. Remember that this problem is looking for "less than" 470, so keep that in mind when looking at the table.
I hope this will help.
Perfect advice. I appreciate the help!
To solve these problems, we will use the properties of the normal distribution and the Z-score formula. The Z-score measures the number of standard deviations a data point is from the mean.
a. To find the probability that a student's score is between 400 and 675, we need to calculate the Z-scores for both values and then find the probability between those values using a Z-table or calculator.
Z1 = (400 - 500) / 100 = -1
Z2 = (675 - 500) / 100 = 1.75
Using a Z-table or calculator, we find that the probability of Z being between -1 and 1.75 is approximately 0.8013.
b. To find the minimum SAT score that corresponds to the top 10%, we need to find the Z-score that corresponds to the top 10% and then convert it back to an actual SAT score.
Using a Z-table or calculator, we find that the Z-score corresponding to the top 10% is approximately 1.28. We can then solve for the actual SAT score by rearranging the Z-score formula:
Z = (X - μ) / σ
1.28 = (X - 500) / 100
Solving for X, we find X ≈ 1.28 * 100 + 500 = 628.
Therefore, the minimum SAT score required to score in the top 10% is 628.
c. To find the probability that the mean score of 49 students is less than 470, we use the Central Limit Theorem, which states that the distribution of sample means approximates a normal distribution with mean μ and standard deviation σ/sqrt(n), where n is the sample size.
In this case, the mean is 500 and the standard deviation is 100. We need to find the Z-score for 470 using the formula:
Z = (X - μ) / (σ / sqrt(n))
Z = (470 - 500) / (100 / sqrt(49))
Simplifying, we have:
Z = -30 / (100 / 7)
Z = -2.1
Using a Z-table or calculator, we find that the probability of Z being less than -2.1 is approximately 0.0179.
Therefore, the probability that the mean score of 49 students is less than 470 is approximately 0.0179.
To find the answers to these questions, you need to use the properties of the normal distribution and the Z-score formula.
a. To find the probability that a randomly selected student's score is between 400 and 675, you need to calculate the area under the normal curve between these two scores. First, convert these scores to Z-scores by subtracting the mean and dividing by the standard deviation:
Z1 = (400 - 500) / 100 = -1
Z2 = (675 - 500) / 100 = 1.75
Next, you can use a Z-score table or a calculator to find the probabilities associated with these Z-scores. Looking up the Z-scores in a table or using a calculator, you'll find that the probability associated with Z1 is 0.1587, and the probability associated with Z2 is 0.9599.
To find the probability between these two scores, subtract the smaller probability from the larger probability:
P(400 ≤ X ≤ 675) = P(Z1 ≤ Z ≤ Z2) = P(Z ≤ Z2) - P(Z ≤ Z1)
= 0.9599 - 0.1587 = 0.8013
So, the probability that a randomly selected student's score on the English part of the SAT is between 400 and 675 is 0.8013.
b. To find the minimum SAT score that a student can receive to score in the top 10%, you need to find the Z-score corresponding to the top 10% of the distribution.
The Z-score corresponding to the top 10% is found by looking up the area of 0.10 (1 - 0.10 = 0.90) in the Z-table. The corresponding Z-score is approximately 1.28.
Now you can use the Z-score formula to find the raw score:
Z = (X - Mean) / Standard deviation
1.28 = (X - 500) / 100
Solving for X:
X - 500 = 1.28 * 100
X - 500 = 128
X = 500 + 128 = 628
So, the minimum SAT score a student can receive to score in the top 10% is 628.
c. To find the probability that the mean score for 49 sampled students is less than 470, you need to use the Central Limit Theorem. The Central Limit Theorem states that the distribution of sample means will be approximately normal as the sample size increases.
Here, you are given the population mean (500), standard deviation (100), and sample size (49). To find the probability, you need to convert the sample mean to a Z-score using the formula:
Z = (X - Mean) / (Standard deviation / √n)
Z = (470 - 500) / (100 / √49)
Z = -30 / (100 / 7)
Z = -30 / 14.14
Z ≈ -2.12
Using a Z-table or calculator, you can find that the probability associated with a Z-score of -2.12 is approximately 0.0179.
Therefore, the probability that the mean score for the 49 sampled students is less than 470 is 0.0179.