Question

Given a normal distribution of the lengths of wing-spans of a certain butterfly has a mean of 20 mm and standard deviation 3.5 mm. Find, ROUNDING YOUR ANSWERS TO 4 DECIMAL PLACES. A) What wing-span has an area of 0.1005 on it left side under the standard normal curve? B) What wing-span has an area of 0.0480 on it right side under the standard normal curve?

Answer 1

Answer:

a) 15.52mm

b) 25.8275mm

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X, or the area to the left side of the curve. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X, that is, the area to the right side of the curve.

A) What wing-span has an area of 0.1005 on it left side under the standard normal curve?

This is X when Z has a pvalue of 0.1005. So it is X when Z = -1.28.

$Z = \frac{X - \mu}{\sigma}$

$-1.28 = \frac{X - 20}{3.5}$

$X - 20 = -1.28*3.5$

$X = 15.52$

This wing-span is 15.52mm.

B) What wing-span has an area of 0.0480 on it right side under the standard normal curve?

This is X when Z has a pvalue of 1 - 0.0480 = 0.9520. So this is X when Z = 1.665.

$Z = \frac{X - \mu}{\sigma}$

$1.665 = \frac{X - 20}{3.5}$

$X - 20 = 1.665*3.5$

$X = 25.8275$

This wing-span is 25.8275mm.