Question
A group of 100 students took a math exam, and the scores
were normally distributed with a mean of 75 and a standard deviation of 10. What is the percentage of students who scored between 60 and 90 on the exam?Solution
Since the scores are normally distributed, we can use the z-score formula to standardize the scores and find the percentage of students who scored between 60 and 90. First, we need to find the z-score for a score of 60 and a score of 90. The z-score formula is: z = (x - μ) / σ where x is the score, μ is the mean, and σ is the standard deviation. For x = 60, we have: z = (60 - 75) / 10 = -1.5 For x = 90, we have: z = (90 - 75) / 10 = 1.5 Now we can use a z-score table or a calculator to find the percentage of students who scored between -1.5 and 1.5 standard deviations from the mean. This percentage represents the percentage of students who scored between 60 and 90 on the exam. Using a z-score table, we find that the percentage of students who scored between -1.5 and 1.5 is approximately 86.6%. Therefore, about 86.6% of the students scored between 60 and 90 on the exam.
(512) (2/3) × √64 ÷ (512) (1/3) = (64) (?/2) ÷ (2)6Â
212.3 × 4414.7 × 4623.4 × 4845.85 = 462?
139 + 323 – √169 + ? = 450
The alarms of two alarm clocks sound at regular intervals of 72 seconds and 80 seconds. If they beep together for the first time at 6:00 am, at what tim...
(22 + √3364)/(? + 4) = 10
- Determine the value of following expression:
[{(148 + 32) ÷ 9}% x 1350] + 19 Simplify: 60 ÷ 5 × 3 + 8 × (7 − 4)
Simplify the given expression:-
33 + (6.25) % of 1600 = ? + 2 X 55
- What will come in the place of question mark (?) in the given expression?
[{(224 + 14 × 23) – 187} × (672 ÷ 28 – ?)] = 1795