Question

The length of a Dachshund is normally distributed with a mean of $15$ inch, and a standard deviation of $3.5$ inch. What length of range would occur $75\%$ of the time?

Answer 1

We have given that length is normally distributed with the known values of mean and standard deviation.
The $Z$ score for the normal distribution curve is given as
$Z = \dfrac{{X - \mu }}{\sigma }$

Complete step by step solution:
We have given that the length of Dachshund is normally distributed with a mean of $15$ inch, and a standard deviation of $3.5$ inch. From given information, we have $\mu = 15$ and $\sigma = 3.5$
Now we have to find the length of range for $75\%$ of the time, so we have to find the range for
$\dfrac{{75\% }}{2}$ of time, which is equal to
$\Rightarrow \dfrac{{75}}{{2 \times 100}}$
$\Rightarrow 0.375$

Now looking at the normal distribution table, the corresponding $Z$ value is $1.15$. Means area under the curve for $Z = 1.15$ is equal to $\approx 0.375$ .
We have to consider both positive and negative values of $Z$ as the normal distribution curve is symmetric about the origin so, $Z = \pm 1.15$

Now considering $Z = \pm 1.15$, applying the formula of normal distribution to calculate $Z$ score, we get
$\pm 1.15 = \dfrac{{X - 15}}{{3.5}}$
Step 4: Now simplifying for the value of $X$ , we get
$\Rightarrow X - 15 = \pm 1.15 \times 3.5$
$\Rightarrow X = 15 \pm 4.025$
Considering positive sign we get $X = 19.025$ and considering negative sign, we get $X = 10.975$ so the range of length is $10.975$ to $19.025$ or $\left( {10.975,19.025} \right)$.

Note: Normal distribution curve is a bell shaped curve and symmetric about origin, so consider both positive and negative values of $Z$ as the area under the curve for both the values is the same.
Standard deviation is the square root of variance.