How to calculate percent from mean and standard deviation

Question

The daily energy intakes of children follow a normal distribution with a mean of 1123 kcal and a standard deviation of 214 kcal. The recommended minimum daily intake is 1700 kcal. What percent of these children receive at least 1700 kcal? 

Answer 1

It sounds like the mean and standard deviation are of the population, so

µ=1123 

σ=214

In this situation, numerous studies would have been conducted to approximate the population parameters, resulting in the decision to deem these values the population parameters. 

To solve this problem, we need to construct a standardized normal distribution, which has µ=0 and σ=1. Since we know the population standard deviation, we would be using a z test. The formula for z is

z=(X-µ)/σ

 

One way to think about standardization is that we are simply finding the amount by which the observed data point is different from the mean (deviation) and then putting it in terms of the standard deviation. If I wanted to convert 12 feet into yards, I would divide the 12 ft by 3 ft, and it would be 4 yards. It's the same with dividing by the standard deviation to determine "what is this data point in terms of standard deviations away from the mean." Given that the observation in which we are interested is 1700,

z=(1700-1123)/214

z=577/214

z=2.69626168=2.70

 

This means that the observation 1700 is 2.7 σ away from the mean. The question asks about the percentage, i.e. proportion, of observations fall above the given observation. Since the space under the normal curve represents 100% of the possible values, finding the proportion requires finding the area under the curve. We want to know the percentage of children that fall above z=2.70, so we must determine the area under the curve between 2.7 and infinity. To find this p value, we could either use a calculator or a z table. Using the latter, we see that the answer is .0035. Thus, .35% of children in this population receive at least 1700 kcal.