Given a mean of 60 and a standard divation of 7, what proportion of scores is between 45 and 78? What is the formula for determining this proportion?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.
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To determine the proportion of scores between 45 and 78, we need to first calculate the z-scores for these values and then use the z-table (normal distribution table). The formula for determining the proportion is as follows:
Proportion = P(Z ≤ Z2) - P(Z ≤ Z1)
Where:
- Z1 is the z-score of the lower value (45 in this case),
- Z2 is the z-score of the higher value (78 in this case),
- P(Z ≤ Z1) is the cumulative probability up to the z-score Z1,
- P(Z ≤ Z2) is the cumulative probability up to the z-score Z2.
To calculate the z-scores, we can use the formula:
Z = (X - μ) / σ
Where:
- X is the value,
- μ is the mean,
- σ is the standard deviation.
In this case, the mean (μ) is 60, and the standard deviation (σ) is 7. Plugging these values into the formula, we can calculate the z-scores for the lower and higher values.
Z1 = (45 - 60) / 7
Z2 = (78 - 60) / 7
Once we have the z-scores, we can use the z-table to look up the corresponding probabilities. Subtracting the cumulative probabilities gives us the proportion of scores between 45 and 78.
To determine the proportion of scores between two values, based on the mean and standard deviation, we use the concept of the Standard Normal Distribution. The Standard Normal Distribution is a bell-shaped curve with a mean of 0 and a standard deviation of 1.
To find the proportion of scores between two values, we need to convert these values into z-scores, which are a measure of how many standard deviations a particular value is away from the mean.
The formula to convert a value into a z-score is:
z = (x - μ) / σ
Where:
z is the z-score,
x is the value of interest,
μ is the mean, and
σ is the standard deviation.
Using this formula, we can determine the z-scores for the values 45 and 78, given a mean of 60 and a standard deviation of 7.
For 45:
z = (45 - 60) / 7 = -2.14 (approximately)
For 78:
z = (78 - 60) / 7 = 2.57 (approximately)
Once we have the z-scores, we can look up the corresponding proportions in the Standard Normal Distribution table or use statistical software.
To find the proportion of scores between these two values, we subtract the cumulative proportion corresponding to the lower z-score from the cumulative proportion corresponding to the higher z-score.
Let's denote the proportion of scores between 45 and 78 as P(45 < x < 78).
P(45 < x < 78) = P(z1 < Z < z2)
Where:
z1 is the z-score for 45,
z2 is the z-score for 78,
Z represents the standard normal distribution.
Finally, we can refer to a standard normal distribution table or use statistical software to look up the proportions or calculate them using the z-scores. For example, the result could be P(45 < x < 78) = 0.822, indicating that approximately 82.2% of scores fall between 45 and 78.