The mean of a set of credit scores is [tex]$\mu=690$[/tex] and [tex]$\sigma=14$[/tex]. Which statement must be true about [tex]$z _{694}$[/tex] ?
A. [tex]$z_{694}$[/tex] is within 1 standard deviation of the mean.
B. [tex]$z_{694}$[/tex] is between 1 and 2 standard deviations of the mean.
C. [tex]$z_{694}$[/tex] is between 2 and 3 standard deviations of the mean.
D. [tex]$z_{694}$[/tex] is more than 3 standard deviations of the mean.
Answer (2)
Calculate the z-score: z = 14 694 − 690 = 14 4 ≈ 0.2857 .
Compare the z-score to the given intervals.
Since 0.2857 < 1 , z 694 is within 1 standard deviation of the mean.
The correct statement is: z 694 is within 1 standard deviation of the mean.
Explanation
Understand the problem We are given that the mean of a set of credit scores is μ = 690 and the standard deviation is σ = 14 . We want to determine which statement is true about z 694 .
Calculate the z-score First, we need to calculate the z-score for x = 694 using the formula: z = σ x − μ where x = 694 , μ = 690 , and σ = 14 .
Substitute the values Plugging in the values, we get: z = 14 694 − 690 = 14 4 = 7 2 ≈ 0.2857
Compare the z-score to the intervals Now we need to determine which of the given statements is true about the calculated z-score, z ≈ 0.2857 .
z 694 is within 1 standard deviation of the mean. This means ∣ z ∣ < 1 . Since 0.2857 < 1 , this statement is true.
z 694 is between 1 and 2 standard deviations of the mean. This means 1 < ∣ z ∣ < 2 . Since 0.2857 is not between 1 and 2, this statement is false.
z 694 is between 2 and 3 standard deviations of the mean. This means 2 < ∣ z ∣ < 3 . Since 0.2857 is not between 2 and 3, this statement is false.
z 694 is more than 3 standard deviations of the mean. This means 3"> ∣ z ∣ > 3 . Since 0.2857 is not greater than 3, this statement is false.
Final Answer Therefore, the correct statement is that z 694 is within 1 standard deviation of the mean.
Examples
Understanding z-scores is useful in many real-world scenarios. For example, in finance, you can use z-scores to assess how far a stock's return is from its average return. In healthcare, z-scores can help determine if a patient's lab results are within a normal range. In education, z-scores can be used to compare a student's performance to the class average. The z-score helps standardize data, making it easier to compare values from different distributions. In this case, we found that a credit score of 694 is less than one standard deviation away from the mean credit score.
The z-score for the credit score of 694 is approximately 0.2857, indicating it is within 1 standard deviation of the mean. Therefore, the correct answer is option A: z 694 is within 1 standard deviation of the mean.
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