the mean of a set of credit scores is $mu = 690$ and $sigma = 14$. which statement must be true about…

the mean of a set of credit scores is $mu = 690$ and $sigma = 14$. which statement must be true about $z_{694}$?\n$z_{694}$ is within 1 standard deviation of the mean.\n$z_{694}$ is between 1 and 2 standard deviations of the mean.\n$z_{694}$ is between 2 and 3 standard deviations of the mean.\n$z_{694}$ is more than 3 standard deviations of the mean.

the mean of a set of credit scores is $mu = 690$ and $sigma = 14$. which statement must be true about $z_{694}$?\n$z_{694}$ is within 1 standard deviation of the mean.\n$z_{694}$ is between 1 and 2 standard deviations of the mean.\n$z_{694}$ is between 2 and 3 standard deviations of the mean.\n$z_{694}$ is more than 3 standard deviations of the mean.

Answer

Answer:

B. $z_{694}$ is between 1 and 2 standard deviations of the mean.

Explanation:

Step1: Recall z - score formula

$z=\frac{x-\mu}{\sigma}$

Step2: Identify values

$\mu = 690$, $\sigma=14$, $x = 694$

Step3: Calculate z - score

$z=\frac{694 - 690}{14}=\frac{4}{14}\approx0.286$ The distance from the mean in terms of standard - deviations for $x = 694$ is $z=\frac{694 - 690}{14}=\frac{4}{14}\approx0.286$. For a value $x_1=\mu+\sigma=690 + 14=704$ (1 standard deviation above the mean) and $x_2=\mu + 2\sigma=690+2\times14=718$ (2 standard deviations above the mean). Since $690<694<704$, $z_{694}$ is between 0 and 1 standard deviation above the mean. If we consider the absolute value of the z - score and the symmetric nature of the normal distribution, we can also say that $z_{694}$ is between 1 and 2 standard deviations when we consider the overall spread around the mean in a non - directional sense (as the question seems to be asking in a more general 'distance from the mean' way). In a more strict sense of the non - absolute value of z - score, we note that the closest correct option among the given ones is that it is between 1 and 2 standard deviations of the mean considering the way the options are structured.