Chapter 3.1, Problem 36E

a.

To determine

To find: The proportion of male fruit flies that are having thorax length above 1 mm.

Answer to Problem 36E

The probability is 0.008.

Explanation of Solution

Given:

  $\begin{array}{l}\text{Population}\text{mean }(\mu )=0.80\\ \text{Standard}\text{deviation}\left(\sigma \right)=0.08\end{array}$

Calculation:

The proportion of male fruit flies that are having thorax length above 1 mm. can be calculated as:

  $\begin{array}{c}P\left(x>1\right)=P\left(\frac{x-\mu }{\sigma }>\frac{1-\mu }{\sigma }\right)\\ =P\left(Z>\frac{1-0.80}{0.08}\right)\\ =P\left(Z>2.50\right)\\ =0.0062\end{array}$

Therefore, the required proportion is 0.0062.

b.

To determine

To find: The ${30}^{\text{th}}$ percentile.

Answer to Problem 36E

The percentile is 0.758.

Explanation of Solution

From standard normal table, the z-value at 0.30 probability is -0.5244

The ${30}^{\text{th}}$ percentile can be calculated as:

  $\begin{array}{c}x=\mu +z\times \sigma \\ =0.80+\left(-0.5244\right)\times 0.08\\ =0.758\end{array}$

Thus, the required percentile is 0.758.