The mass of blue whales are normally distributed with a mean of $$ 170,000 \mathrm{~kg} $$. What is the standard deviation if $$ 34 \% $$ of blue whales have a mass between $$ 170,000 \mathrm{~kg} $$ and $$ 178,000 \mathrm{~kg} $$ ?

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1. Let the mass of a blue whale be $$X$$ with distribution $$X \sim N(170,\!000, \sigma^2)$$.

2. Since $$34\%$$ of blue whales have a mass between $$170,\!000 \, \text{kg}$$ and $$178,\!000 \, \text{kg}$$, we have:
   $$
   P(170,\!000 \leq X \leq 178,\!000) = 0.34.
   $$

3. Standardize the variable by letting $$Z = \frac{X - 170,\!000}{\sigma}$$. Since $$170,\!000$$ is the mean, this simplifies to:
   $$
   P(0 \leq Z \leq \frac{178,\!000 - 170,\!000}{\sigma}) = P\left(0 \leq Z \leq \frac{8,\!000}{\sigma}\right) = 0.34.
   $$

4. In the standard normal distribution, the probability from the mean (i.e. $$Z = 0$$) to a positive value $$z$$ is given by:
   $$
   P(0 \leq Z \leq z) = \Phi(z) - 0.5,
   $$
   where $$\Phi(z)$$ is the cumulative distribution function (CDF).

5. Setting up the equation:
   $$
   \Phi\left(\frac{8,\!000}{\sigma}\right) - 0.5 = 0.34.
   $$

6. Solve for $$\Phi\left(\frac{8,\!000}{\sigma}\right)$$:
   $$
   \Phi\left(\frac{8,\!000}{\sigma}\right) = 0.84.
   $$

7. From the standard normal table, $$\Phi(z) = 0.84$$ corresponds approximately to:
   $$
   z \approx 1.
   $$
   Hence,
   $$
   \frac{8,\!000}{\sigma} \approx 1.
   $$

8. Solving for $$\sigma$$:
   $$
   \sigma \approx 8,\!000 \, \text{kg}.
   $$

Thus, the standard deviation is $$\sigma = 8,\!000 \, \text{kg}$$.
The standard deviation of the blue whale masses is 8,000 kg.

The mass of blue whales are normally distributed with a mean of \( 170,000 \mathrm{~kg} \). What is the standard deviation if \( 34 \% \) of blue whales have a mass between \( 170,000 \mathrm{~kg} \) and \( 178,000 \mathrm{~kg} \) ?

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