Answer :
To answer the question about what percent of scores is less than a score of 65, we would typically need to know the distribution of the scores. Commonly, test scores are assumed to follow a normal distribution. If this is the case, we would need additional information: the mean (average score) and the standard deviation (which measures variability around the mean) of the distribution of scores.
Assuming that we have a normal distribution, here's how we would calculate the percentage of scores that are less than 65 using the mean and standard deviation:
1. First, we would need to find the z-score for a score of 65. The z-score tells us how many standard deviations a data point (score of 65, in this case) is from the mean. The formula to calculate the z-score is:
[tex]\( z = \frac{(X - \mu)}{\sigma} \)[/tex]
where [tex]\( X \)[/tex] is the score of interest (65 in this case), [tex]\( \mu \)[/tex] is the mean, and [tex]\( \sigma \)[/tex] is the standard deviation.
2. Once we have the z-score, we would then refer to the standard normal distribution table (or use a statistical calculator or statistical software) to find the proportion of scores that fall below that z-score. This is because the standard normal distribution has a mean of 0 and a standard deviation of 1, and z-scores correspond to this standard dataset.
3. The standard normal distribution table gives us the area under the curve to the left of the z-score. This represents the percentage of scores that are less than our score of interest.
Let's pretend for a moment that we have a mean score ([tex]\( \mu \)[/tex]) of 100 and a standard deviation ([tex]\( \sigma \)[/tex]) of 15 for the distribution of scores we are considering. Here's how the calculation would go:
Step 1: Calculate the z-score.
[tex]\( z = \frac{(X - \mu)}{\sigma} \)[/tex]
[tex]\( z = \frac{(65 - 100)}{15} \)[/tex]
[tex]\( z = \frac{-35}{15} \)[/tex]
[tex]\( z \approx -2.33 \)[/tex]
Step 2: Use the z-score to find the percentage.
Looking at the z-table or using a statistical calculator for a z-score of approximately -2.33, we find that about 1% of scores are less than a score of 65. This means that if the scores are normally distributed with a mean of 100 and a standard deviation of 15, roughly 1% of the scores are less than 65.
Without the actual values for the mean and standard deviation, we cannot provide an accurate percentage. You would follow the steps above using the mean and standard deviation for your specific distribution of scores to get the correct percentage. If the distribution is not normal, the method to find the percentage may differ.
Assuming that we have a normal distribution, here's how we would calculate the percentage of scores that are less than 65 using the mean and standard deviation:
1. First, we would need to find the z-score for a score of 65. The z-score tells us how many standard deviations a data point (score of 65, in this case) is from the mean. The formula to calculate the z-score is:
[tex]\( z = \frac{(X - \mu)}{\sigma} \)[/tex]
where [tex]\( X \)[/tex] is the score of interest (65 in this case), [tex]\( \mu \)[/tex] is the mean, and [tex]\( \sigma \)[/tex] is the standard deviation.
2. Once we have the z-score, we would then refer to the standard normal distribution table (or use a statistical calculator or statistical software) to find the proportion of scores that fall below that z-score. This is because the standard normal distribution has a mean of 0 and a standard deviation of 1, and z-scores correspond to this standard dataset.
3. The standard normal distribution table gives us the area under the curve to the left of the z-score. This represents the percentage of scores that are less than our score of interest.
Let's pretend for a moment that we have a mean score ([tex]\( \mu \)[/tex]) of 100 and a standard deviation ([tex]\( \sigma \)[/tex]) of 15 for the distribution of scores we are considering. Here's how the calculation would go:
Step 1: Calculate the z-score.
[tex]\( z = \frac{(X - \mu)}{\sigma} \)[/tex]
[tex]\( z = \frac{(65 - 100)}{15} \)[/tex]
[tex]\( z = \frac{-35}{15} \)[/tex]
[tex]\( z \approx -2.33 \)[/tex]
Step 2: Use the z-score to find the percentage.
Looking at the z-table or using a statistical calculator for a z-score of approximately -2.33, we find that about 1% of scores are less than a score of 65. This means that if the scores are normally distributed with a mean of 100 and a standard deviation of 15, roughly 1% of the scores are less than 65.
Without the actual values for the mean and standard deviation, we cannot provide an accurate percentage. You would follow the steps above using the mean and standard deviation for your specific distribution of scores to get the correct percentage. If the distribution is not normal, the method to find the percentage may differ.
