Question

Laboratory experiment shows that the life of the average butterfly is normally distributed with a mean of 18.8 days and a standard deviation of 2 days.

Find the probability that a butterfly will live between 12.04 and 18.38 days.

a) 0.4164

b) 0.4203

c) 0.5828

d) 0.3893

e) 0.4202

f) None of the above.

Answer 1

Answer:

a) 0.4164

Step-by-step explanation:

Mean lifespan (μ) = 18.8 days

Standard deviation (σ) = 2 days

For any given lifespan 'X', the z-score is:

$z=\frac{X- \mu}{\sigma} \\z=\frac{X- 18.8}{2}$

For X=12.04 days:

$z=\frac{12.04 - 18.8}{2} \\z=-3.38$

A z-score of -3.38 falls in the 0.036-th percentile of a normal distribution.

For X=18.38 days:

$z=\frac{18.38 - 18.8}{2} \\z=-0.21$

A z-score of -3.38 falls in the 41.68th percentile of a normal distribution.

The probability that a butterfly lives between 12.04 and 18.38 days is

$P=41.68-0.036\\P=41.64\%\ or\ 0.4164$