Question

Duck #1 lays eggs whose weights are normally distributed with a mean of 70gramsand a standard deviation of 6 grams.Duck #2 lays eggs whose weights are also normally distributed with a mean of 65 grams and a standard deviation of 5 grams. If an egg is randomly chosen from each duck, what is the probability that Duck #2’s egg weighs more than Duck #1’s egg?

Answer 1

Answer:

26.11% probability that Duck #2’s egg weighs more than Duck #1’s egg

Step-by-step explanation:

When the distribution is normal, we use 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.

If an egg is randomly chosen from each duck, what is the probability that Duck #2’s egg weighs more than Duck #1’s egg?

Duck #2 egg will weigh more if the subtraction of duck's 2 egg by duck's 1 egg is larger than 0.

When we subtract normal distributions, the mean is the subtraction of the means. So

$\mu = 65 - 70 = -5$

The standard deviation is the square root of the sum of the variances. So

$\sigma = \sqrt{6^2+5^2} = \sqrt{61} = 7.81$

Now, we have to find 1 subtracted by the pvalue of Z when X = 0. So

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

$Z = \frac{0 - (-5)}{7.81}$

$Z = 0.64$

$Z = 0.64$ has a pvalue of 0.7389

1 - 0.7389 = 0.2611

26.11% probability that Duck #2’s egg weighs more than Duck #1’s egg