Question

Weights of adult males are normally distributed with a mean of u = 85.5 kg and a standard deviation of o = 17.7 kg. If a sample yields x = 80 kg, how many standard deviations is this away from the mean?

Answer 1

An individual weighing 80 kg is approximately 0.31 standard deviations below the mean weight of adult males, which is 85.5 kg with a standard deviation of 17.7 kg.

Step-by-step explanation:

When determining how many standard deviations a data point x is from the mean μ, you calculate the z-score using the formula z = (x - μ) / σ, where x is the sample mean, μ is the population mean, and σ is the population standard deviation. In the case where the mean weight (μ) of adult males is 85.5 kg and the standard deviation (σ) is 17.7 kg, an individual weighing 80 kg would be:

z = (80 kg - 85.5 kg) / 17.7 kg = -5.5 kg / 17.7 kg ≈ -0.31

This means the individual is approximately 0.31 standard deviations below the mean weight.

Since the z-score is negative, it means that the sample value is below the mean. The absolute value of the z-score tells us how many standard deviations away from the mean the sample value is. In this case, the sample value is approximately 0.3104 standard deviations below the mean.