Question

When dragons on planet Pern lay eggs, the eggs are either green or yellow. The biologists have observed over the years that 32% of the eggs are yellow, and the rest green. Next spring the lead scientist has permission to randomly select 59 of the dragon eggs to incubate. Consider all the possible samples of 59 dragon eggs.

What is the usual number of yellow eggs in samples of 59 eggs? (Give answers as SENSIBLE whole numbers.)


minimum usual number of yellow eggs =



maximum usual number of yellow eggs =

Answer 1

Usual number of yellow eggs: 19 (rounded to whole number)

Minimum usual number of yellow eggs: 16

Maximum usual number of yellow eggs: 22

Step-by-step explanation:

Expected proportion of yellow eggs: We know 32% of eggs are yellow, so the expected proportion of yellow eggs in a sample is 0.32.

Expected number of yellow eggs: Multiply the expected proportion by the sample size: 0.32 * 59 eggs ≈ 18.88 eggs.

Usual number of yellow eggs: Since whole eggs are counted, we round 18.88 to the nearest whole number, which is 19. This represents the usual (typical) number of yellow eggs in a sample of 59.

Minimum and maximum usual range: Due to random sampling, the actual number of yellow eggs can vary within a reasonable range. We can estimate this range using the standard deviation:

Standard deviation for binomial distribution = sqrt(p * (1-p) * n)

In this case, it's sqrt(0.32 * 0.68 * 59) ≈ 3.42.

Add and subtract the standard deviation from the expected number: 19 ± 3.42.

Therefore, the minimum usual number of yellow eggs is 16 (19 - 3.42) and the maximum usual number is 22 (19 + 3.42).

These values represent the typical range of yellow eggs we can expect in a sample of 59 dragon eggs based on the population proportion of 32%.

Answer 2

Answer and Step-by-step explanation:

The computation is shown below

Given that as per the question

n = 59

p = 0.32

Therefore mean is

`=\mu`

`= np`

`= 59 * 0.32`

= 18.88

Now the standard deviation i.e
`\sigma`

`= √(np(1-p))`

`= √(18.88(1 - 0.32))`

= 3.58

Now the minimum usual number of yellow eggs is

`= \mu - 2* \sigma`

`= 18.88-2 * 3.58`

= 12 yellow eggs

And, the maximum number is

`= \mu + 2* \sigma`

`= 18.88 + 2 * 3.58`

= 26 yellow eggs