For 25 customer ages with mean 31.68 and standard deviation 9.694, the minimum age is 11 and the maximum is 47. Standardize both.

Step-by-step solution

1. 1

   Write the standardising formula. A $z$-score expresses a value as a number of standard deviations away from the mean:

   $z=\frac{x-\bar x}{s}$

   The subtraction must come first; dividing before subtracting is the usual slip.

2. 2

   Standardise the minimum.

   $z_{\min}=\frac{11-31.68}{9.694}=\frac{-20.68}{9.694}=-2.133$

   The negative sign says the youngest customer sits below the mean — by a little over two standard deviations.

3. 3

   Standardise the maximum.

   $z_{\max}=\frac{47-31.68}{9.694}=\frac{15.32}{9.694}=1.580$

4. 4

   State both results.

   $\boxed{z_{\min}=-2.133,\qquad z_{\max}=1.580}$

5. 5

   Interpret the pair. Both magnitudes are under $3$, so by the common rule of thumb neither extreme is an unusual outlier. But they are noticeably unequal: the minimum is $2.13$ standard deviations below the mean while the maximum is only $1.58$ above it, so the low tail stretches further — evidence of a left-skewed age distribution.

6. 6

   Check the arithmetic by reversing it. $\bar x+z_{\min}s=31.68-2.133(9.694)=31.68-20.68=11.00$ ✓ and $31.68+1.580(9.694)=31.68+15.32=47.00$ ✓, recovering the original ages.