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#### karush

##### Well-known member

- Jan 31, 2012

- 2,998

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 $\approx 43\%$