"Runzheimer International publishes data on overseas business travel costs. They report that the
average per diem total for a business traveller in Paris, France, is $349. Suppose the shape of the
distribution of the per diem costs of a business traveller to Paris is unknown, but that 51% of the
per diem figures are between $320 and $378. What is the value of the standard deviation?" Runzheimer International publishes data on overseas business travel costs. They report that the average per diem total for a business traveller in Paris, France, is $349. Suppose the shape of the distribution of the per diem costs of a business traveller to Paris is unknown, but that 51% of the per diem figures are between $320 and $378. What is the value of the standard deviation?
mean=349
1-1/k^2*100=0.51
k=1.00
320-419/SD=1
SD=-99 is wrong?
What did I do wrong/supposed to do?
Let $X$ represent the per diem cost of a randomly selected business traveler. From Chebychev's inequality (taking $k=10/7$) we have $$P\Bigg(\Bigg|\frac{X-349}{{\sigma}}\Bigg|\geq \frac{10}{7}\Bigg)\leq \bigg(\frac{7}{10}\bigg)^2 $$ Notice $$\Bigg|\frac{X-349}{{\sigma}}\Bigg|\geq \frac{10}{7} \iff X\in\bigg(-\infty,349-\frac{10 \sigma}{7}\bigg]\cup\bigg[349+\frac{10 \sigma}{7},\infty\bigg)$$ We need to find $\sigma$ so that $$349-\frac{10 \sigma}{7}=320$$ $$349+\frac{10 \sigma}{7}=378$$ Solving these for $\sigma$ gives $\sigma =\frac{203}{10}$