Life Spans of Fruit Flies The life spans of a species of

Step 1 of 2:

Given the life spans of a species of fruit fly have a bell-shaped distribution, with the mean of 33 days and the standard deviation of 4 days.

Here \(\mu=33\) and \(\sigma=4\).

Our goal is :

a). We need to find the z-score that corresponds to each life span.

b). We need to find the percentile that corresponds to each life span.

a). Given the life spans of 3 randomly selected fruit flies are 34 days, 30 days, and 42 days.

Now we have to find a z-score.

The z-score formula is

\(\mathrm{z}=\frac{x-\mu}{\sigma}\)

The life span of 34 days is

\(\mathrm{z}=\frac{x-\mu}{\sigma}\)

We know that \(\mu=33\) and \(\sigma=4\).

\(\begin{array}{l} z=\frac{34-33}{4} \\ z=\frac{1}{4} \\ z=0.25 \end{array}\)

Therefore, the z-score is 0.25.

The life span of 30 days is

\(\mathrm{z}=\frac{\frac{x-\mu}{\sigma}}{\sigma}\)

We know that \(\mu=33\) and \(\sigma=4\).

\(\begin{array}{l} z=\frac{30-33}{4} \\ z=\frac{-3}{4} \\ z=-0.75 \end{array}\)

Therefore, the z-score is -0.75.

The life span of 42 days is

\(\mathrm{z}=\frac{x-\mu}{\sigma}\)

We know that \(\mu=33\) and \(\sigma=4\).

\(\begin{array}{l} z=\frac{42-33}{4} \\ z=\frac{9}{4} \\ z=2.25 \end{array}\)

Therefore, the z-score is 2.25.

Hence the lifespan of 42 days is unusual due to a rather large-score.