QUESTION IMAGE
Question
- the weight of oranges growing in an orchard is normally distributed with a mean weight of 5.5 oz. and a standard deviation of 0.5 oz. using the empirical rule, what percentage of the oranges from the orchard weigh between 5 oz. and 6 oz.?
options: 68%, 65%, 95%, 99.7%
Step1: Identify mean and standard deviation
Mean ($\mu$) = 5.5 oz, Standard Deviation ($\sigma$) = 0.5 oz.
Step2: Find how many standard deviations from mean
For 5 oz: $5 = 5.5 - 0.5 = \mu - \sigma$
For 6 oz: $6 = 5.5 + 0.5 = \mu + \sigma$
Step3: Apply Empirical Rule
Empirical Rule states that for normal distribution, about 68% of data lies within $\mu \pm \sigma$.
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68% (corresponding to the option with "68%")