In a population of exam scores, a score of X=88 corresponds to z= +2.00 and a score of X=79 corresponds to
z= -1.00. Find the mean and standard deviation for the population.
Mean is 2 SD below 88 and one above 79.
88 - 79 = 9 Therefore mean is 1/3 of that distance above 79 and 2/3 below 88.
Z = (x - μ)/SD
Substitute the values for Z, x and μ, and solve for the SD.
I hope this helps. Thanks for asking.
To find the mean and standard deviation for the population, we need to use the z-score formula and the property of z-scores.
The formula for converting a raw score (X) to a z-score (z) is:
z = (X - mean) / standard deviation
From the given information, we have:
X1 = 88, z1 = 2.00
X2 = 79, z2 = -1.00
We can set up two equations using the z-score formula and the given data:
For X1 = 88:
2.00 = (88 - mean) / standard deviation (equation 1)
For X2 = 79:
-1.00 = (79 - mean) / standard deviation (equation 2)
To solve these equations, we can use a method called "elimination". Multiply equation 2 by 2 and add it to equation 1:
2 * (-1.00) = 2 * (79 - mean) / standard deviation
-2.00 = (158 - 2mean) / standard deviation
2.00 = (88 - mean) / standard deviation
Adding these two equations together:
-2.00 + 2.00 = (158 - 2mean) / standard deviation + (88 - mean) / standard deviation
0 = (246 - 3mean) / standard deviation
Multiplying both sides of the equation by the standard deviation:
0 * standard deviation = 246 - 3mean
0 = 246 - 3mean
Rearranging the equation, we get:
3mean = 246
mean = 246 / 3 = 82
Now, let's substitute the value of mean (82) into one of the original equations (equation 1):
2.00 = (88 - 82) / standard deviation
2.00 = 6 / standard deviation
Multiplying both sides by the standard deviation:
2.00 * standard deviation = 6
standard deviation = 6 / 2.00 = 3
Therefore, the mean of the population is 82 and the standard deviation is 3.
To find the mean and standard deviation for the population, we will use the concept of standard scores, also known as z-scores.
A standard score (z-score) is a measure of how many standard deviations a particular value is from the mean of a distribution. It helps us understand how each individual score relates to the overall distribution.
Given that a score of X=88 corresponds to z=+2.00 and a score of X=79 corresponds to z=-1.00, we can use these values to calculate the mean and standard deviation.
Step 1: Finding the mean
The z-score formula is given by:
z = (X - mean) / standard deviation
For the score X=88 and z=+2.00:
2.00 = (88 - mean) / standard deviation
For the score X=79 and z=-1.00:
-1.00 = (79 - mean) / standard deviation
Step 2: Solving the simultaneous equations
We have two equations:
2.00 = (88 - mean) / standard deviation (Equation 1)
-1.00 = (79 - mean) / standard deviation (Equation 2)
We can solve this system of equations to find the values of mean and standard deviation.
By substituting mean = 88 - 2.00 * standard deviation from Equation 1 into Equation 2, we get:
-1.00 = (79 - (88 - 2.00 * standard deviation)) / standard deviation
Simplifying the equation:
-1.00 = (79 - 88 + 2.00 * standard deviation) / standard deviation
Multiplying both sides by standard deviation:
-1.00 * standard deviation = 79 - 88 + 2.00 * standard deviation
Rearranging the equation:
-1.00 * standard deviation - 2.00 * standard deviation = 79 - 88
Simplifying further:
-3.00 * standard deviation = -9
Dividing both sides by -3.00:
standard deviation = -9 / -3.00 = 3.00
Now, we can substitute the value of standard deviation back into Equation 1 to find the mean:
2.00 = (88 - mean) / 3.00
Cross-multiplying:
2.00 * 3.00 = 88 - mean
Simplifying:
6.00 = 88 - mean
Rearranging the equation:
mean = 88 - 6.00
mean = 82.00
Therefore, the mean of the population is 82.00 and the standard deviation is 3.00.