Question

The adrenaline concentrations of all sparrows in a county are normally distributed with a mean of 0.3 gram and a standard deviation of 0.02 gram. What percentage of the sparrows, taken 10 at a time, have mean concentration between 0.28 and 0.33 gram? Round your answer to the nearest integer. Do not include the percentage symbol in your answer.

Answer 1

Answer:

77.45% of the sparrows, taken 10 at a time, have mean concentration between 0.28 and 0.33 gram.

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:

The adrenaline concentrations of all sparrows in a county are normally distributed with a mean of 0.3 gram and a standard deviation of 0.02 gram. This means that $\mu = 0.3, \sigma = 0.02$.

What percentage of the sparrows, taken 10 at a time, have mean concentration between 0.28 and 0.33 gram?

This percentage is the pvalue of Z when $X = 0.33$ subtracted by the pvalue of Z when $X = 0.28$. So

X = 33

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

$Z = \frac{0.33 - 0.3}{0.02}$

$Z = 1.5$ has a pvalue of 0.9332

X = 28

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

$\frac{0.28-0.30}{0.02}$

$Z = -1$ has a pvalue of  0.1587

This means that 0.9332 - 0.1587 = 0.7745 = 77.45% of the sparrows, taken 10 at a time, have mean concentration between 0.28 and 0.33 gram.