How do they get this answer?
What proportion of the students scoring between 292 and 310 in Professor Zangs?
ZANG: MEAN: 295 STANDARD DEVIATION: 13
Answer: .4659 or 47%
they used a z-score table to find the portion of the population between 3/13 s.d below the mean, and 15/13 s.d. above the mean
3/13 = .231 s.d. ... z-score below mean
15/13 = 1.154 s.d. ... z-score above mean
take the z-scores into the table, and it will give the portion of the population that lies in the range
add the two portions (above and below the mean)
To find the proportion of students scoring between 292 and 310 in Professor Zangs, you need to follow these steps:
1. Calculate the z-scores for the lower and upper bounds.
- 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 the lower bound (292):
z_lower = (292 - 295) / 13 = -0.23
For the upper bound (310):
z_upper = (310 - 295) / 13 = 1.15
2. Look up the corresponding cumulative probabilities for the z-scores.
- You can use a standard normal distribution table or a statistical calculator to find the cumulative probabilities.
For the lower bound (z_lower = -0.23), the cumulative probability is approximately 0.4090 or 40.90%.
For the upper bound (z_upper = 1.15), the cumulative probability is approximately 0.8749 or 87.49%.
3. Calculate the proportion of students scoring between the two bounds.
Proportion = cumulative probability (upper bound) - cumulative probability (lower bound)
Proportion = 0.8749 - 0.4090 = 0.4659
So, the proportion of students scoring between 292 and 310 in Professor Zangs is approximately 0.4659, which is equivalent to 46.59% or approximately 47%.