The average time a student spends studying just prior to a test is 36 minutes with a standard deviation of 5 minutes. This distribution approximates a normal curve. Find the percent of students who study just prior to the test for. a) more than 40 minutes b) between 35 and 45 minutes
ncdf(40,1000,36,5)=.212 a)
ncdf(35,45,36,5)=.543 b)
its been a while so im not 100% sure if its right but i think it is :)
dont forget to draw the normal curves on your assignment my teachers alwyas made me haha
Z = (score-mean)/SD
Calculate the various Z scores.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to these Z scores
To find the percent of students who study just prior to the test for more than 40 minutes, we need to calculate the z-score and use the standard normal distribution table.
a) More than 40 minutes:
First, calculate the z-score using the formula: z = (X - μ) / σ
Where:
X = the value we want to find the percentage for (40 minutes)
μ = the mean (average) time spent studying (36 minutes)
σ = the standard deviation (5 minutes)
z = (40 - 36) / 5
z = 4 / 5
z = 0.8
Now, we can use the standard normal distribution table to find the area to the right of z = 0.8 (since we want the percentage of students who study more than 40 minutes).
From the table, the area to the left of z = 0.8 is 0.7881.
Since we want the area to the right of z = 0.8, subtract the left area from 1:
Area to the right = 1 - 0.7881 = 0.2119
So, approximately 21.19% of students study for more than 40 minutes just prior to the test.
b) Between 35 and 45 minutes:
To find the percentage of students who study between 35 and 45 minutes, we need to find the area to the left of both z-scores and subtract the smaller area from the larger area.
First, calculate the z-scores for 35 and 45 minutes using the same formula as before.
For 35 minutes:
z = (35 - 36) / 5
z = -1 / 5
z = -0.2
For 45 minutes:
z = (45 - 36) / 5
z = 9 / 5
z = 1.8
Now, use the standard normal distribution table to find the area to the left of each z-score.
From the table, the area to the left of z = -0.2 is 0.4207, and the area to the left of z = 1.8 is 0.9641.
Finally, subtract the smaller area from the larger area:
Area between 35 and 45 minutes = 0.9641 - 0.4207 = 0.5434
So, approximately 54.34% of students study between 35 and 45 minutes just prior to the test.