- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 3,187

1149

The lifespan of a particular species of insect is normally distributed with a mean of $57$ hours and a standard deviation of $4.4$ hours.

$90\%$ of the insects die after $t$ hours.

Represent this information on a standard normal curve diagram,

indicating clearly the area representing $90\%$.

$.9$ on the z-table $\displaystyle \approx z=1.29$. so from

Find the value of t.

$\displaystyle\frac{t-57}{4.4}=1.29$ thus $t\approx 63s$

The lifespan of a particular species of insect is normally distributed with a mean of $57$ hours and a standard deviation of $4.4$ hours.

$90\%$ of the insects die after $t$ hours.

Represent this information on a standard normal curve diagram,

indicating clearly the area representing $90\%$.

$.9$ on the z-table $\displaystyle \approx z=1.29$. so from

**W|A**Find the value of t.

$\displaystyle\frac{t-57}{4.4}=1.29$ thus $t\approx 63s$

Last edited: