the five number summary of the distribution of scores on a statistics exam is

0 26 31 36 50
316 students took the exam. the histogram of all 316 scores was approximately normal. Thus the variance of test scores must be about
I know about subtracting 26-31 over the standard dev but how do you find the z-score? Then squaring the standard dev. to find variance

So you can assume that the mean is equal to the median because the data is equally distributed. Also, we know that 25% of the data lies between the given median and either Q1 or Q3. So, using z values, the median/mean=.5, and if you're using Q1, find a z value closest to .25 so that the area in between is roughly 25%. You can then use the standardizing equation to solve for the unknown standard deviation. Set it equal to the z value for Q1 (or you can use Q3 with another z value). So on top, you would have 26-31 and divide by s. Solve for s and then square it because variance is s^2 - you should get 55

55 verified

To find the z-score, you need the mean and the standard deviation of the distribution. In this case, we can estimate the mean by finding the average of the minimum and maximum values in the five-number summary:

Mean = (0 + 50) / 2 = 25

To calculate the standard deviation, you can use the formula for the range:

Range = Maximum - Minimum = 50 - 0 = 50

Since the distribution is approximately normal, we assume that the range spans approximately 6 standard deviations. Therefore, the estimated standard deviation would be:

Standard Deviation ≈ Range / 6 = 50 / 6 ≈ 8.33

Now, we can calculate the z-score for each value using the formula:

z-score = (x - mean) / standard deviation

For example, for the value 26:

z-score = (26 - 25) / 8.33 ≈ 0.12

Next, you need to square the standard deviation to find the variance:

Variance = Standard Deviation^2 = 8.33^2 ≈ 69.33

Therefore, the estimated variance of the test scores is approximately 69.33.

To find the variance of the test scores, we can use the five-number summary and the number of students who took the exam.

Before calculating the variance, let's review some key concepts:

1. Z-score: A z-score measures the number of standard deviations a data point is from the mean. To find the z-score for a specific value, we use the formula: (value - mean) / standard deviation.

2. Variance: Variance is a measure of how spread out the data is. It is the average of the squared deviations from the mean.

Now, let's calculate the variance step by step:

Step 1: Calculate the mean.
To find the mean, we add up all the scores and divide by the total number of scores:
mean = (0 + 26 + 31 + 36 + 50) / 5 = 143 / 5 = 28.6

Step 2: Calculate the standard deviation.
To find the standard deviation, we need the deviations of each score from the mean. Here are the deviations:
(0 - 28.6), (26 - 28.6), (31 - 28.6), (36 - 28.6), (50 - 28.6)

The sum of the squared deviations is:
(0 - 28.6)^2 + (26 - 28.6)^2 + (31 - 28.6)^2 + (36 - 28.6)^2 + (50 - 28.6)^2
= 814.64

To find the standard deviation, we divide this sum by (n-1), where n is the number of scores:
standard deviation = sqrt(814.64 / (5-1)) = sqrt(814.64 / 4) ≈ sqrt(203.66) ≈ 14.27

Step 3: Calculate the variance.
To find the variance, we square the standard deviation:
variance = standard deviation^2 = 14.27^2 ≈ 203.66

So, the variance of the test scores is approximately 203.66.