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

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

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 = (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 = (X - \mu)/(\sigma)`

`Z = (0.33 - 0.3)/(0.02)`

`Z = 1.5` has a pvalue of 0.9332

X = 28

`Z = (X - \mu)/(\sigma)`

`(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.