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A distribution of values is normal with a mean of 211.3 and a standard deviation of 55.7.

Find the probability that a randomly selected value is between 38.6 and 300.4.

P(38.6 < x < 300.4) =

Answer :

The probability of a randomly selected value being between 38.6 and 300.4 in a normal distribution is approximately 95.33%.

To find the probability that a randomly selected value is between 38.6 and 300.4 in a normal distribution with mean 211.3 and standard deviation 55.7, we need to standardize the values and use the z-score formula.

The z-score for 38.6 is calculated as (38.6 - 211.3) / 55.7 = -3.069,

and the z-score for 300.4 is calculated as (300.4 - 211.3) / 55.7 = 1.603.

Using a standard normal distribution table or a calculator,

we can find the probabilities corresponding to these z-scores.

P(-3.069 < z < 1.603) is approximately 0.9533, or 95.33%.

The probability of a value falling between 38.6 and 300.4 in a normal distribution (mean 211.3, standard deviation 55.7) is about 95.33%.

This is found by standardizing the values using z-scores: -3.069 for 38.6 and 1.603 for 300.4.

Using a standard normal distribution table, the probability between these z-scores is calculated. This indicates a high likelihood that a randomly selected value from the distribution lies within the specified range, illustrating the typical behavior of data clustering around the mean in a normal distribution.

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