Question

Determine the $90th$ percentile of the distribution, which is $N(65,25)$.

Short answer

The 90th percentile of the distribution $N(65,25)$ is approximately 71.408.

Step-by-step solution

Interpret the Given Distribution

A normal distribution is given as $N(65,25)$. This means that the distribution has a mean ($\mu$) of 65 and a variance (${\sigma }^2$) of 25. Therefore, the standard deviation ($\sigma$) is the square root of the variance, which is 5.

Convert the Percentile to a z-score

The z-score given as the 90th percentile corresponds to an area of 0.9 under the standard normal curve. Looking this value up in a z-table or using a calculator gives us a z-score of approximately 1.2816.

Use the z-score Formula to Determine the Raw Score

The z-score formula is given by $z=(X-\mu )/\sigma$. We rearrange this formula to solve for X: $X=z\sigma +\mu$. Substituting the given z-score, standard deviation, and mean, we find $X=1.2816\ast 5+65=71.408$.

Interpret the Result

The 90th percentile of the distribution $N(65,25)$ is approximately 71.408. This means that 90% of the values in this distribution are less than 71.408.

Key concepts

Understanding Normal Distribution

When studying statistics, one fundamental concept is the normal distribution, often referred to as the 'bell curve'. It is a symmetrical, bell-shaped distribution where most of the observations cluster around the central peak, and the probabilities for values further away from the mean taper off equally in both directions. It's essential to recognize that the normal distribution assumes that data spreads around a central value (the mean) with a certain standard deviation, which measures the data's variability.

In the context of our exercise, the normal distribution is expressed as $N(65,25)$. This notation provides two critical pieces of information. The first number, 65, is the mean $\mu$ of the distribution, indicating where the peak of the bell is centered. The second number, 25, is the variance ${\sigma }^2$, which provides a sense of how spread out the distribution is. Since the variance is the square of the standard deviation $\sigma$, we take the square root of 25 to find that the standard deviation is 5 for this particular set of data. This indicates how much individual data points are expected to deviate from the mean.

The Role of Z-Score in Distribution

The z-score is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations. It is a way of standardizing scores on the same scale by subtracting the mean and then dividing this difference by the standard deviation.

Mathematically, the z-score for an observation $X$ is calculated using the formula $z=(X-\mu )/\sigma$. This formula helps us to determine how far away, in standard deviations, a certain value is from the mean. If the z-score is 0, it means the value is exactly at the mean. A positive z-score indicates the value is above the mean, whereas a negative z-score signifies a value below the mean.

In our exercise, we used the z-score to find the 90th percentile. Percentiles are used to understand and interpret data by indicating the relative standing of an observation within a dataset. So, the 90th percentile means that 90% of the data falls below this point. We found the corresponding z-score that matches the 90th percentile by using z-tables or statistical software, which indicated a z-score of approximately 1.2816. This tells us that the 90th percentile is 1.2816 standard deviations above the mean.

Standard Deviation as a Measure of Spread

Standard deviation is a statistic that measures the dispersion of a dataset relative to its mean and is calculated as the square root of the variance. In simpler terms, it tells us on average how far the individual data points are from the mean of the data set. A low standard deviation means that the data points tend to be very close to the mean, while a high standard deviation indicates that the data is spread out over a large range of values.

Returning to our example, we had a variance of 25. By taking the square root of the variance, we found a standard deviation of 5. This value is crucial as it helped us to interpret the z-score and understand the distribution of the data. With the standard deviation, we could convert the calculated z-score back to the actual value in the context of the given normal distribution. This conversion from z-score back to raw score is what led us to conclude that the 90th percentile of the distribution $N(65,25)$ is approximately 71.408. Therefore, the standard deviation is not just a measure of spread but also a key component in accurately locating percentiles within a normal distribution.