# [SOLVED]-aux04 Species of insect is normally distributed

#### karush

##### Well-known member
just seeing how I did on this one 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.

this is the normal distribution with $\mu = 57$ and $\sigma = 4.4$

View attachment 1026
tried to standardize this by $\frac{55-75}{4.4}=-0.45$ and $\frac{60-57}{4.4}=0.68$

with $\mu = 0$ and $\sigma = 1$ and $P(-0.45 < x < 0.68)$
which hopefully looks like the given graph on the right below

View attachment 1027View attachment 1028

(a) What are the values of $a$ and $b$
from the standard $\frac{x-\mu}{\sigma} a=-0.45$ and $b=0.68$

(b) Find the probability that the lifespan of an insect of this species is more than $55$ hours

$P(55 < X)$ from $z$ score $0.45$ then $0.1736 + .5 = .6736$ or $\approx 67\%$

View attachment 1029

(b) Find the probability that the lifespan of an insect of this species is between $55$ and $60$ hours

$0.2517+0.1736=0.4253$ or $\approx 43\%$

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#### MarkFL

Staff member
Re: species of insect is normally distributed

Using rational numbers instead of decimal approximations:

a) $$\displaystyle a=-\frac{5}{11},\,b=\frac{15}{22}$$

Using numeric integration for comparison:

b) $$\displaystyle P(55<X)\approx0.675282$$

c) $$\displaystyle P(55<X<60)\approx0.427605$$

I would say you did them correctly.

#### karush

##### Well-known member
Re: species of insect is normally distributed

well that is encouraging. there still is one more question on this but I will post it tomorrow. #### karush

##### Well-known member
90% of the insects die after t hours.

(i) Represent this information on a standard normal curve diagram indicating clearly the area representing $90\%$.

(ii) Find the value of $t$.

does this mean $P(X < t)$, also should $60 < t$ I don't see where this $90\%$ is supposed to be since P(55 < X < 60) was [FONT=MathJax_Main]0.427605[/FONT]