91Ó°ÊÓ

We are given that sample has $45$adult females.

Mean $\overline{x}=14.7$

Standard deviation $s=3.1$

According to chebyshev's Rule

$$\begin{gathered} \overline{x}=14.7 \\ \overline{x}-s=11.6 \\ \overline{x}-2s=8.5 \\ \overline{x}-3s=5.4 \\ \overline{x}+s=17.8 \\ \overline{x}+2s=20.9 \\ \overline{x}+3s=24.0 \end{gathered}$$

Required graph is

We are given that sample has $45$adult females

Mean $\overline{x}=14.7$

Standard deviation $s=3.1$

According to Chebyshev's rule ,if

$k=2$, then at least $75\%$of the adult females in the sample within two standard deviations to either side of the mean.

Now,

$45×0.75≈34$

Now, two standard deviations to either side of the mean is from

$8.5$to $20.9$milligram.

Interpretation- At least $34$out of $45$adult females in the sample have recommended dietary allowances of iron between $8.5mg$and $20.9mg$

We are given that sample has $45$adult females.

Mean$\overline{x}=14.7$

Standard deviation $s=3.1$

According to Chebyshev's rule ,if

$k=3$, then at least $89\%$of the adult females are within two standard deviations to either side of the mean.

Now,

$45×0.89≈40$

Now, two standard deviations to either side of the mean is from

$5.4$to $24.0$.

Interpretation- At least $40$out of $45$adult females in the sample have recommended dietary allowance of iron between $5.4mg$ and $24.0mg$