Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Step-by-step explanation:
Understanding the Problem
We’re dealing with a normal distribution of test scores. We know the average score (mean) and how spread out the scores are (standard deviation). We want to find a specific test score © where 81% of the students scored lower than that score.
Using the Z-Score
Convert to a Z-score: We need to convert our desired probability (81%) into a standard z-score. A z-score tells us how many standard deviations away from the mean a particular value is. You can use a Z-table or a calculator to find the z-score that corresponds to a cumulative probability of 0.81 (representing 81%). You’ll find that the z-score is approximately 0.88.
Apply the Z-score formula: The z-score formula is:
z = (X - μ) / σ
Where:
z = z-score
X = the raw score we’re trying to find (in this case, C)
μ = the mean (74)
σ = the standard deviation (11)
Solve for C: Plug in the values and solve for C:
0.88 = (C - 74) / 11
Multiply both sides by 11: 9.68 = C - 74
Add 74 to both sides: C ≈ 83.7
Answer
The test score C, rounded to one decimal place, is 83.7. This means that approximately 81% of the students scored lower than 83.7
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.