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: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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]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 [tex]\mu = 0.3, \sigma = 0.02[/tex].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 [tex]X = 0.33[/tex] subtracted by the pvalue of Z when [tex]X = 0.28[/tex]. SoX = 33[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{0.33 - 0.3}{0.02}[/tex][tex]Z = 1.5[/tex] has a pvalue of 0.9332X = 28[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]\frac{0.28-0.30}{0.02}[/tex][tex]Z = -1[/tex] has a pvalue of  0.1587This 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.