T2.5. The average yearly snowfall in Chillyville is Normally distributed with a mean of $55$ inches. If the snowfall in Chillyville exceeds $60$inches in $15\%$ of the years, what is the standard deviation?
(a) $4.83$ inches
(d) $8.93$ inches
(b) $5.18$ inches
(c) The standard deviation
(c) $6.04$ inches cannot be computed from the given information.

The average yearly snowfall in Chillyville is normally distributed with a mean of $55$inches. And the snowfall in Chillyville exceeds$60$inches in $15\%$ of the years.

The probability that the snowfall $X$exceeds $60$inches is $15\%$. And the Mean$=55$ inches.
Then,

$$\begin{gathered} P(X>60)=15\% \\ =0.15 \end{gathered}$$

In total, the probability needs to be$1$. So, the probability that $X$does not exceed $60$is then $1$(or $100\%$) decreased by $P(X>60)$

$$\begin{gathered} P(X\le 60)=1-P(X>60) \\ =1-0.15 \\ =0.85 \end{gathered}$$

The probability closest to $0.85$in the table is $0.8508$, which is given in the row $1.0$and column $.04$, thus the probability then corresponds with the z-score of $1.0+.04=1.04$.

$z=1.04$

The standardized $z$-score is the value $x$decreased by the mean, divided by the standard deviation$\sigma$.
$z=\frac{x-\mu }{\sigma }$
Multiply by$\sigma$:
$z\sigma =x-\mu$
Divide by$z$:
$\sigma =\frac{x-\mu }{z}$

Evaluate:

$$\begin{gathered} \sigma =\frac{x-\mu }{z} \\ =\frac{60-55}{1.04} \\ =\frac{5}{1.04} \\ \approx 4.8077 \end{gathered}$$
Hence, the standard deviation is approximately $4.8077$inches, which is closest to $4.83$inches among the given answer options.