On a small farm, the weights of eggs that young hens lay are normally distributed with a mean weight of 51.3 grams and a standard deviation of 4.8 grams. Using the 68-95-99.7 rule, about what percent of eggs weigh between 46.5g and 65.7g.

83.85% of the population is between 46.5g and 65.7g

Explanation:

the 68-95-99.7 rule states that the probability that a random variable (X=Weights of the eggs) is found inside 1 standard deviation, 2 standard deviations and 3 standard deviations from the mean is 68%, 95%, 99.7% respectively

then calling Z to

Z = (X-μ) / σ

where μ = mean, σ = standard deviation, then

for X₁=46.5g

Z₁ = (X₁-μ) / σ = (46.5g - 51.3g) / 4.8 g = - 1

1 standard deviation from both sides (±1) = 68%

then since the normal distribution is symmetrical

1 standard deviation from one side (±1) = 68%/2 = 34%

for X₂=46.5g

Z₂ = (X₂-μ) / σ = (65.7 g - 51.3g) / 4.8 g = + 3

3 standard deviation from both sides (±1) = 99.7%

then since the normal distribution is symmetrical

3 standard deviation from one side (±1) = 99.7%/2 = 49.85%

then

between - 1 standard deviation from the mean and + 3 standard deviations from the mean there is = 34% + 49.85% = 83.85% of the population (since the regions do not overlap each other)