Unfortunately, I can't draw a histogram for you here as this platform does not support visual content creation. However, I can guide you through the process and help you calculate the mean and the standard deviation using the assumed mean method.
a. To draw a histogram:
1. Choose class intervals that will cover the range of your scores (here, from 50 to 96).
2. Tally the number of scores that fall within each class interval.
3. Determine the frequency of each class.
4. On the x-axis, mark the class intervals. On the y-axis, mark the frequency.
5. For each class interval, draw a bar that has a height corresponding to its frequency.
b. Calculations using an assumed mean:
i. Mean of the distribution:
Using an assumed mean \( A = 67 \):
We'll set up a table:
| Class Interval | Midpoint (x) | Frequency (f) | \( x - A \) | \( f \cdot (x - A) \) |
|-----------------|--------------|---------------|-------------|----------------------|
| 50-54 | 52 | 5 | -15 | -75 |
| 55-59 | 57 | 12 | -10 | -120 |
| 60-64 | 62 | 16 | -5 | -80 |
| 65-69 | 67 | 22 | 0 | 0 |
| 70-74 | 72 | 18 | 5 | 90 |
| 75-79 | 77 | 10 | 10 | 100 |
| 80-84 | 82 | 6 | 15 | 90 |
| 85-89 | 87 | 3 | 20 | 60 |
| 90-96 | 93 | 8 | 26 | 208 |
| Total | | 100 | | |
Now, sum up the \( f \cdot (x - A) \) column and divide by total frequency (100):
\( f \cdot (x - A) \) total sum = -75 - 120 - 80 + 0 + 90 + 100 + 90 + 60 + 208 = 273
Mean (\( \overline{x} \)) = \( A + \frac{\sum f \cdot (x - A)}{N} \)
Mean \( \overline{x} \) = 67 + (273 / 100) = 67 + 2.73 = 69.73
ii. Standard deviation:
To calculate the standard deviation, we need to find each class's squared deviation (\( (x - A)^2 \)) and the frequency times the squared deviation (\( f \cdot (x - A)^2 \)):
| Class Interval | Midpoint (x) | \( x - A \) | \( (x - A)^2 \) | \( f \cdot (x - A)^2 \) |
|-----------------|--------------|-------------|----------------|-------------------------|
| 50-54 | 52 | -15 | 225 | 1125 |
| 55-59 | 57 | -10 | 100 | 1200 |
| 60-64 | 62 | -5 | 25 | 400 |
| 65-69 | 67 | 0 | 0 | 0 |
| 70-74 | 72 | 5 | 25 | 450 |
| 75-79 | 77 | 10 | 100 | 1000 |
| 80-84 | 82 | 15 | 225 | 1350 |
| 85-89 | 87 | 20 | 400 | 1200 |
| 90-96 | 93 | 26 | 676 | 5408 |
| Total | | | | |
Sum of \( f \cdot (x - A)^2 \) = 1125 + 1200 + 400 + 0 + 450 + 1000 + 1350 + 1200 + 5408 = 12133
Standard deviation \( \sigma \) = \( \sqrt{\frac{\sum f \cdot (x - A)^2}{N} - (\frac{\sum f \cdot (x - A)}{N})^2} \)
\( \sigma \) = \( \sqrt{\frac{12133}{100} - (2.73)^2} \)
\( \sigma \) = \( \sqrt{121.33 - 7.4529} \)
\( \sigma \) = \( \sqrt{113.8771} \)
\( \sigma \) ≈ 10.7
iii. Probability of not having less than 74:
We calculate the z-score for 74:
Z = \( \frac{X - \overline{x}}{\sigma} \)
Z = \( \frac{74 - 69.73}{10.7} \)
Z ≈ \( \frac{4.27}{10.7} \)
Z ≈ 0.399
In a normal distribution, a z-score of 0.399 would correspond to a probability. We need to look at standard normal distribution tables or use a calculator to find the probability that corresponds to Z = 0.399. It would be around 65.5%. This is the probability of a student scoring less than 74.
So, the probability that a student did not have less than 74 is:
1 - Probability(Z < 0.399)
≈ 1 - 0.655
≈ 0.345 or 34.5%
Please note that the standard deviation and the probability calculation assume that the distribution is normal or close to normal, which may not be the case. You'd need to check the distribution shape using the histogram you draw.
An examination was conducted for 100 students and the results are given in the below
50,51,52,53,54,55,55,55,56,56,56,57,57,57,58,58,58,59,59,59, 60,60,60,60,61,61,61,61,62,62,62,62,63,63,63,63,64,64,64,64, 65,65,65,65,65,65,66,66,66,66,66,66,67,67,67,67,67,67,68,68, 68,68,68,68,69,69,69,69,70,70,71,71,71,72,72,72,73,73,73,74, 75,75,76,76,77,77,78,78,79,80,80,81,82,84,85,87,89,90,94,96.
a. Draw a histogram for the data.
b. Using an assumed mean of 67, calculate, correct to one decimal place.
i. the mean of the distribution
ii. the standard deviation of the distribution
iii. if a student is selected at random, what is the probability that he/she did not have less than 74.
1 answer