To find the standard deviation of the scores, we need to use the information given about the mean, the percentage above 90, and the assumption of an approximately normal distribution.
Let's break down the problem step by step:
1. The mean of the scores is given as 78.4. This means that the average score is 78.4.
2. We are told that 7 percent of the scores are above 90. To find the standard deviation, we need to determine the z-score corresponding to this percentage.
The z-score represents the number of standard deviations a data point is from the mean in a normal distribution. Since the distribution is assumed to be normal, we can use the z-score formula to calculate it:
z = (X - μ) / σ
where X is the score, μ is the mean, and σ is the standard deviation.
In this case, we want to find the z-score for the value 90. Rearranging the formula, we have:
σ = (X - μ) / z
Plugging in the values, we have:
σ = (90 - 78.4) / z
To find the z-score corresponding to 7 percent above 90, we can use a standard normal distribution table or a calculator. The z-score that corresponds to 7 percent is approximately 1.88.
Therefore, the formula becomes:
σ = (90 - 78.4) / 1.88
≈ 6.17
The standard deviation is approximately 6.17 when rounded to the nearest tenth.
So, the standard deviation of the scores is approximately 6.2 when rounded to the nearest tenth.