The test grades at a large school have an approximately normal distribution with a mean of 50. What is the standard deviation of the data so that 80% of the students are within 12 points (above or below) the mean? Draw a sketch to help reason the procedure and be clever. A.5.875 B.14.5 C.9.375 D.Cannot be determined from the given information. E.10.375

Answer:

Step-by-step explanation:

The 80% of students within 12 points of the mean are captured by the shaded area, which is two standard deviations away from the mean on either side.

We know that the normal distribution curve is symmetrical, so the two shaded areas have equal areas. Therefore, each shaded area has an area of 40%.

We can use the inverse normal cumulative distribution function (also known as the probit function) to find the z-scores corresponding to the 40th percentile and the 60th percentile. These z-scores are both equal to 1.28.

The z-score is defined as the number of standard deviations away from the mean:

z = (x - μ) / σ

where:

x is the observed value

μ is the mean

σ is the standard deviation

Therefore, the standard deviation is equal to:

σ = (x - μ) / z

From the sketch above, we can see that the x-value corresponding to the 40th percentile is 38 and the x-value corresponding to the 60th percentile is 62.

Substituting these values into the equation above, we can solve for the standard deviation:

σ = (62 - 50) / 1.28 = 9.375

Therefore, the standard deviation of the data so that 80% of the students are within 12 points of the mean is 9.375.

Cleverness:

To solve this problem cleverly, we can use the fact that the normal distribution curve is symmetrical. This means that we can find the standard deviation by calculating the distance between the mean and the 40th percentile, and then multiplying by 2. This is because the 40th percentile is the same distance away from the mean as the 60th percentile.

Final answer:

In a normal distribution with a mean of 50, if 80% of the grades are within 12 points of the mean, the standard deviation can be calculated as approximately 9.375 (option C). We obtained this result by equating the z-score that represents 80% of the data to the ratio of the score range (12 points) to the standard deviation, and solving for the standard deviation.

Explanation:

To determine the standard deviation given 80% of the scores are within 12 points from the mean, we need to understand the properties of a normal distribution curve. In a normal distribution, about 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and almost all (99.7%) falls within three standard deviations. Now, when we say 80% of the data falls within 12 points of the mean, we are looking at an area of the curve that is equivalent to +/- one standard deviation around the mean.

In a standard normal distribution, an area of approximately 80% correlates to a z-score of about 1.28 (using a z-table or z-score calculator). The z-score is calculated as (X - μ) / σ, where X is the score, μ is the mean and σ is the standard deviation.

In our case, we have X = ±12 (since it's 12 points above and below the mean), and μ = 0 (since we're looking at the difference from the mean). So, 1.28 = 12 / σ. Solving for σ, we get that σ = 12 / 1.28 ≈ 9.375. So the standard deviation, which provides the answer to your question, is 9.375 (option C).

Learn more about Standard Deviation here:

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