Mark owns siberian husky sled dogs. he knows from data collected over the years that the weight of the dogs is a normal distribution. they have a mean weight of 52.5 lbs and a standard deviation of 2.4 lbs. what percentage of his dogs would you expect to have a weight between 47.7 lbs and 54.9 lbs?

Calculate the z-score for the given data points in the item using the equation,
 
                             z-score = (x - μ) / σ

where x is the data point, μ is the mean, and σ is the standard deviation.

Substituting,
             (47.7)     z-score = (47.7 - 52.5)/2.4 = -2

This translates to a percentile of 2.28%.

              (54.9)      z-score = (54.9 - 52.5)/2.4 = 1

This translates to a percentile of 84.13%. 

Then, subtract the calculate percentiles to give us the final answer of 81.85%. 

Thus, 81.85% of the Siberian Husky sled dogs are expected to weigh between 47.7 and 54.9 lbs. 

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