Answer :
To find the percentage of values that lie below a certain value in a normal distribution, we need to calculate the z-score for that value and then refer to the standard normal distribution (which has a mean of 0 and a standard deviation of 1) to find the corresponding percentile.
The z-score is calculated using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where:
- [tex]\( X \)[/tex] is the value for which we want to find the percentage,
- [tex]\( \mu \)[/tex] is the mean of the distribution, and
- [tex]\( \sigma \)[/tex] is the standard deviation of the distribution.
Given:
- [tex]\( \mu = 24 \)[/tex] (mean)
- [tex]\( \sigma = 7 \)[/tex] (standard deviation)
- [tex]\( X = 3 \)[/tex]
Now let's calculate the z-score for the value 3:
[tex]\[ z = \frac{3 - 24}{7} \][/tex]
[tex]\[ z = \frac{-21}{7} \][/tex]
[tex]\[ z = -3 \][/tex]
Having the z-score of -3, we can use the empirical rule, also known as the 68-95-99.7 rule, to estimate the percentage of values below this score. However, the empirical rule states that approximately 99.7% of data within a normal distribution falls within three standard deviations from the mean. This means that about 0.3% of data lies outside of three standard deviations from the mean.
Because the z-score of -3 suggests that the value 3 lies three standard deviations below the mean, we can infer that about half of that 0.3%, which would be approximately 0.15%, of values lie below 3.
Therefore, the percentage of values that lie below 3 in our normal distribution is approximately 0.15%. This roughly matches your provided answer, which seems to be correct given the context of the empirical rule. If you require a more precise answer, normally you would use standard normal distribution tables or a computational tool to find the exact percentile.
The z-score is calculated using the formula:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where:
- [tex]\( X \)[/tex] is the value for which we want to find the percentage,
- [tex]\( \mu \)[/tex] is the mean of the distribution, and
- [tex]\( \sigma \)[/tex] is the standard deviation of the distribution.
Given:
- [tex]\( \mu = 24 \)[/tex] (mean)
- [tex]\( \sigma = 7 \)[/tex] (standard deviation)
- [tex]\( X = 3 \)[/tex]
Now let's calculate the z-score for the value 3:
[tex]\[ z = \frac{3 - 24}{7} \][/tex]
[tex]\[ z = \frac{-21}{7} \][/tex]
[tex]\[ z = -3 \][/tex]
Having the z-score of -3, we can use the empirical rule, also known as the 68-95-99.7 rule, to estimate the percentage of values below this score. However, the empirical rule states that approximately 99.7% of data within a normal distribution falls within three standard deviations from the mean. This means that about 0.3% of data lies outside of three standard deviations from the mean.
Because the z-score of -3 suggests that the value 3 lies three standard deviations below the mean, we can infer that about half of that 0.3%, which would be approximately 0.15%, of values lie below 3.
Therefore, the percentage of values that lie below 3 in our normal distribution is approximately 0.15%. This roughly matches your provided answer, which seems to be correct given the context of the empirical rule. If you require a more precise answer, normally you would use standard normal distribution tables or a computational tool to find the exact percentile.
