Range Question!

Question

70 students took an exam. the mean score on the exam is 75.5 and the standard deviation is 2.6 what percent of the scores are in the 72.9 and 83.3 range?

Answer 1

look at a standard deviation chart

one standard deviation includes 68.2% (±34.1%)

two standard deviations includes 95.6%, 68.2+27.2% (±13.6%)

three standard deviations includes 99.8%, 95.6%+4.2% (±2.1%)

standard deviation is 2.6

72.9 to 83.3 is 75.5-2.6=72.9 and 75.5+2.6+2.6+2.6=83.3

so, 68.2% (one standard deviation)+13.6%+2.1%=83.9%

Answer 2

Question: 70 students took an exam. the mean score on the exam is 75.5 and the standard deviation is 2.6 what percent of the scores are in the 72.9 and 83.3 range?

Answer: Let's set up a number line that could be the foundation for modeling the shape and spread of the exam scores. The very middle of it is the mean of 75.5. Subtract 2.6 three times (marking the differences along the way), and then starting again add the mean, add 2.6 three times (marking the sums along the way).

75.5 - 2.6 = 72.9

75.5 - 2.6 - 2.6 = 70.3

75.5 - 2.6 - 2.6 - 2.6 = 67.7

75.5 + 2.6 = 78.1

75.5 + 2.6 + 2.6 = 80.7

75.7 + 2.6 + 2.6 + 2.6 = 83.3

Lay out the numbers in order:

67.7, 70.3, 72.9, 75.5, 78.1, 80.7, 83.3

We're asked about the percent of scores in the 72.9 and 83.3 range.

72.9 is one standard deviation below average, and 83.3 is three standard deviations above average. Use the empirical rule to approach this problem. If 68% of scores fall within one standard deviation of the mean (such as the 72.9) in both directions, then we can look at spread in two halves, each of 34%. I'm only interested in looking to the left to grab that 72.9. That's 34% already. Now I need to grab the 83.3 somehow. The empirical rule states that 99.7% of scores fall within three standard deviations of the mean in both directions, but I only need to look rightwards to grab the 83.3. I'll cut 99.7% in half to get 49.85%.

Now, Add up 34% and 49.85% to get 83.85%. A calculator will do this task more precisely, but going strictly off of the empirical rule gets you quite close.

83.85% of scores fall in the 72.9 and 83.3 range.

Answer 3

Let x be the exam score. The mean of x is 75.5 and the standard deviation is 2.6.

Because of the large sample size (70) we can assume the population is normally distributed and

z = (x - mean)/standard deviation = (x = 75.5/2.6) is normally distributed with mean 0 and standard deviation 1

z is the standardized normal statistic which can be used to determine probabilities from the standard normal probability table.

we are trying to find the probability (which is equal to percentage/100) that (72.9 < x < 83.3)

We first transform the range of the inequality from x to z, so we are finding:

P(72.9 < x < 83.3) = P((72.9-75.5/2.6) < z < (83.3 - 75.5)/2.6) = P(-1 < z < 3).

Since the probabilities in the standard normal probability table are cumulative to the left:

P(-1 < z < 3) = P(z < 3) - P(z < -1) = 0.9987 - 0.1587 = 0.84, which is equal to 84%