Reading placement scores for a particular college are normally distributed with a mean of 72 and a standard deviation of 7.
a) draw a sketch of the reading placement scores for students at this college. be sure to mark the mean and also to mark one and two standard deviations from each side of the mean.
b) 95% of students at this college have a reading placement score between ______ and ______
c) Sammy's reading placement score is 82. What is his z-score?
d) Find the proportion of students at this college who have a reading placement score of at least 82.
e) Elisa's z-score is -1.5. What is her reading placement score?
f) How high does one's reading placement score have to be so that a student is in the 85th percentile?
a. http://en.wikipedia.org/wiki/Standard_deviation
b. 95% = mean ± 1.96 SD
c. Z = (score-mean)/SD
d. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion from the Z score, using equation above.
e. Use same equation.
f. Use same table to find Z score and calculate.
a) To sketch the reading placement scores for students at this college, we need to draw a normal distribution curve.
First, draw a horizontal axis labeled "Reading Placement Scores" with tick marks representing the scores. Then, draw a symmetric bell-shaped curve above the axis. The curve should be centered at the mean, which is 72 in this case. Mark the mean with a vertical line passing through it.
To mark one standard deviation, which is 7 in this case, draw two vertical lines passing through points that are 7 units away from the mean on each side. These lines indicate the boundary for one standard deviation from the mean.
To mark two standard deviations, which is 14 (twice the standard deviation) in this case, draw two more vertical lines passing through points that are 14 units away from the mean on each side. These lines indicate the boundary for two standard deviations from the mean.
b) To find the reading placement scores between which 95% of students fall, we use the concept of z-scores and the empirical rule. We know the mean (72) and standard deviation (7) of the distribution.
The empirical rule states that approximately 68% of the data falls within one standard deviation of the mean, approximately 95% of the data falls within two standard deviations, and approximately 99.7% of the data falls within three standard deviations.
Since we want to find the range that includes 95% of the scores, we take two standard deviations above and below the mean.
The lower boundary:
Mean - (2 * standard deviation) = 72 - (2 * 7) = 58
The upper boundary:
Mean + (2 * standard deviation) = 72 + (2 * 7) = 86
Therefore, 95% of students have reading placement scores between 58 and 86.
c) To find Sammy's z-score for his reading placement score of 82, we can use the formula:
z = (x - mean) / standard deviation,
where x is the reading placement score, mean is the mean of the distribution (72), and the standard deviation is given (7).
Calculating:
z = (82 - 72) / 7 = 1.43
So, Sammy's z-score is 1.43.
d) To find the proportion of students at this college who have a reading placement score of at least 82, we need to find the area under the normal distribution curve to the right of 82.
We can use a standard normal distribution table or a statistical software to find this area. The z-score we just calculated for Sammy is 1.43, which represents the standardized score for 82 in a standard normal distribution.
Using a standard normal distribution table or software, we find that the proportion to the right of 1.43 is approximately 0.9236.
Therefore, approximately 0.9236 (or 92.36%) of students at this college have a reading placement score of at least 82.
e) To find Elisa's reading placement score given her z-score of -1.5, we use the formula:
x = (z * standard deviation) + mean,
where z is the z-score (in this case, -1.5), standard deviation is given (7), and mean is the mean of the distribution (72).
Calculating:
x = (-1.5 * 7) + 72 = 61.5
So, Elisa's reading placement score is 61.5.
f) To determine the reading placement score required to be in the 85th percentile, we need to find the z-score that corresponds to the 85th percentile and then use it to find the reading placement score.
The 85th percentile means that 85% of the data falls below that score. This implies that 15% of the data falls above it.
Using a standard normal distribution table or software, we find the z-score that corresponds to the 85th percentile as approximately 1.04.
To find the reading placement score:
x = (z * standard deviation) + mean,
Calculating:
x = (1.04 * 7) + 72 = 79.28
Therefore, a student's reading placement score needs to be approximately 79.28 to be in the 85th percentile.