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Bond [772]
3 years ago
15

The adrenaline concentrations of all sparrows in a county are normally distributed with a mean of 0.3 gram and a standard deviat

ion 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.
Mathematics
1 answer:
Lorico [155]3 years ago
7 0

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.

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