The mean of Exam 1 is 80%. The variance is 25. How many students of my IS310 class of 200 students have scores between 70% and 90%, assuming that the distribution is bell-shaped?

Question

The mean of Exam 1 is 80%. The variance is 25. How many students of my IS310 class of 200 students have scores between 70% and 90%, assuming that the distribution is bell-shaped?

a.180
b.185
c.190
d.195

I'm not sure how to set up this problem.

Answer 1

Z = (score-mean)/SEm

SEm = SD/√n

SD = √variance

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability for each and the Z score. Insert into above equation.

Answer 2

I'm not allowed to use the chart. I have to calculate it by hand

Answer 3

So I think the way the professor wanted to answer this is knowing the empirical rule because its normal

65%= 1 SD
95%=2 SD
98%=3 SD
all from the mean

Variance of 25 gives us a Standard deviation of 5.

80% is the mean. So moving 2 SD up gives us 90%

Its normal and the empirical rule states that 2 SD is 95%

So you multiply (95%)x 200= 190

Answer 4

To solve this problem, we need to use the concept of the standard deviation and the z-score. The z-score measures how many standard deviations a data point is away from the mean of a distribution.

First, let's find the standard deviation. The variance is given as 25, so the standard deviation (denoted as σ) is the square root of the variance, which is √25 = 5.

Next, we can find the z-scores for the lower and upper boundaries. To find the z-score, we can use the formula:

z = (x - μ) / σ

where x is the score, μ is the mean, and σ is the standard deviation.

For the lower bound, x = 70, μ = 80, and σ = 5. Plugging these values into the formula gives us:

z_lower = (70 - 80) / 5 = -10 / 5 = -2

For the upper bound, x = 90, μ = 80, and σ = 5. Plugging these values into the formula gives us:

z_upper = (90 - 80) / 5 = 10 / 5 = 2

Now, we need to find the area under the bell-shaped distribution curve between these two z-scores. This area represents the proportion of students with scores between 70% and 90%.

Using a z-table or a statistical calculator, we can find that the area between z = -2 and z = 2 is approximately 0.9545. This means that approximately 95.45% of the students have scores between 70% and 90%.

Finally, we can calculate the number of students by multiplying the proportion by the total number of students:

Number of students = Proportion * Total number of students
= 0.9545 * 200
≈ 190

Therefore, the answer is c. 190.