Question

In an albatross colony the wingspan of the birds is approximately normally distributed with standard deviation 0.43 metres. In this colony, 67% of the birds have wingspan less than 3.21 metres. Find the average wingspan of the birds in this colony. Give your answer as a decimal, rounded to two decimal places. Do not include the unit.

Answer 1

Answer:

The answer is 3.02.

Step-by-step explanation:

Problems of normally distributed samples can be 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\sigma = 0.43$

67% of the birds have wingspan less than 3.21 metres.

This means that the pvalue of Z when X = 3.21 is 0.67. So when X = 3.21, Z = 0.44.

We have to find $\mu$

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

$0.44 = \frac{3.21 - \mu}{0.43}$

$3.21 - \mu = 0.44*0.43$

Multiplying by -1

$\mu - 3.21 = -0.44*0.43$

$\mu = 3.02$

The answer is 3.02.