Question

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.

Answer 1

Answer:

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)