For normally distributed data, we can use the z-score formula:
[tex]\begin{gathered} Z=\frac{x-\mu}{\sigma} \\ \text{Where } \\ z\text{ is the z-score} \\ \mu\text{ is the mean} \\ \sigma\text{ is the standard deviation} \end{gathered}[/tex]The given information is that the mean is 38 min/day, also 84% of dog owners spend less than 45 min/day walking. This means that P(x<45)=0.84.
By checking in a standard normal table, we found that a probability of 0.84 is given by a z-score = 0.9946
By replacing this value into the z-score formula, we can solve for standard deviation as follows:
[tex]\begin{gathered} 0.9946=\frac{45-38}{\sigma} \\ \sigma=\frac{45-38}{0.9946} \\ \sigma=\frac{7}{0.9946} \\ \sigma=7.03801\approx7.04 \end{gathered}[/tex]To construct the normal distribution curve, you can use the mean and the standard deviation:
In the bottom, replace the mean symbol with its value 38, and standard deviation by 7.04, and you'll get your curve.
Now, to find how long would a dog-owner need to walk in a day to be ranked in the 99th percentile, Start by finding the z-score for P(ZLooking in a standard normal table we found that z-score for 0.99 is 2.33.
Now, replace this value into the z-score formula and solve for x:
[tex]\begin{gathered} 2.33=\frac{x-38}{7.03801} \\ x-38=2.33\times7.03801 \\ x-38=16.3986 \\ x=16.3986+38 \\ x=54.3986\approx54.4 \end{gathered}[/tex]Thus, a dog-owner needs to walk 54.4 min/day to be ranked in the 99th percentile.
