150 students sat exams. The following elements were calculated
ΣFX = 10200
Σ(x-μ)F = 196348
Determine the number of students whose marks were within 2 standard deviations of the mean score
Are you sure that second formula isn't Ó(x-ì)²F? Unless it is, I can't see a way of calculating a standard deviation from the statistics supplied.
On the assumption that the second statistic is Ó(x-ì)²F, the variance ought to be (1/150) times this figure, which is 196348/150 = 1309, so the standard deviation would be 36.2, with a mean of 10200/150=204. So any student with a score of between 204-(2x36.2) and 204+(2x36.2) would be within two standard deviations of the mean.
Having said all that, you asked for the actual number of students within that range - and it seems to me that you don't actually need either of the statistics given to answer that question. If the scores are Normally distributed, then 4.6% of the distribution ought to lie outside two standard deviations (i.e. either in the lower or the upper tail). Out of 150 students that's 7 students - so 143 ought to lie inside plus or minus two standard deviations. (I can't see any way to tell *exactly* how many students lie within that interval, even from the statistics supplied.)
(Sorry - the symbols Ó and ì ought to be a sigma sign and a mu respectively. You evidently can't just copy them from another posting - it doesn't work.)
To determine the number of students whose marks were within 2 standard deviations of the mean score, we need to calculate the mean and the standard deviation of the scores.
Step 1: Calculate the mean score (μ)
The mean score can be calculated using the formula:
μ = ΣFX / ΣF
where ΣFX represents the sum of the products of each score (x) and its frequency (F), and ΣF is the sum of the frequencies.
In this case, ΣFX = 10200 and ΣF = 150.
So, μ = 10200 / 150 = 68.
Step 2: Calculate the standard deviation (σ)
The standard deviation can be calculated using the formula:
σ = √(Σ(x-μ)²F / ΣF)
where Σ(x-μ)²F represents the sum of the squares of the difference between each score (x) and the mean (μ), multiplied by its frequency (F), and ΣF is the sum of the frequencies.
In this case, Σ(x-μ)F = 196348 and ΣF = 150.
So, σ = √(196348 / 150) ≈ √1308.32 ≈ 36.15.
Step 3: Determine the number of students within 2 standard deviations of the mean
To find the number of students whose marks were within 2 standard deviations of the mean score, we need to find the range of scores that fall within ±2σ of the mean score.
The range within 2 standard deviations of the mean can be calculated by:
Lower Range = μ - 2σ
Upper Range = μ + 2σ
In this case, Lower Range = 68 - 2(36.15) ≈ -4.30
and Upper Range = 68 + 2(36.15) ≈ 136.30
Since it is not possible to have negative scores, we ignore the Lower Range and only consider the Upper Range.
Now, to determine the number of students, we need to look at the frequency distribution of scores and count the number of students whose scores fall within the Upper Range.
Note: Without knowing the specific frequency distribution of scores, we cannot provide an exact answer.