The weight of oranges growing in an orchard is normally distributed with a mean weight of ( 7.5 mathrm{oz} ). and a standard deviation of ( 1.5 mathrm{oz} ). Using the empirical rule, determine what interval would represent weights of the middle ( 95 % ) of all oranges from this orchard.

Question

Accepted Answer

Step:1 ：(95%) of the data lies within (pm 2) standard deviations of the mean.Step:2 ：Lower limit: (7.5 - 2(1.5) = 4.5) ozStep:3 ：Upper limit: (7.5 + 2(1.5) = 10.5) oz

Answer

Step: 1 ：(95%) of the data lies within (pm 2) standard deviations of the mean.Step: 2 ：Lower limit: (7.5 - 2(1.5) = 4.5) ozStep: 3 ：Upper limit: (7.5 + 2(1.5) = 10.5) oz

The weight of oranges growing in an orchard is normally distributed with a mean weight of $7.5 oz$ . and a standard deviation of $1.5 oz$ . Using the empirical rule, determine what interval would represent weights of the middle $95 %$ of all oranges from this orchard.

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The weight of oranges growing in an orchard is normally distributed with a mean weight of 7.5oz. and a standard deviation of 1.5oz. Using the empirical rule, determine what interval would represent weights of the middle 95% of all oranges from this orchard.

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The weight of oranges growing in an orchard is normally distributed with a mean weight of $7.5 oz$ . and a standard deviation of $1.5 oz$ . Using the empirical rule, determine what interval would represent weights of the middle $95 %$ of all oranges from this orchard.