Recently, a random sample of 13-18 year olds was asked, "How much do you currently have in savings?" The data in the table re
present the responses to the survey. Approximate the mean and standard deviation amount of savings.
1 answer:
Answer:
μ = 235.38
σ = 234.54
Step-by-step explanation:
Assuming the table is as follows:
This is an example of grouped data, where a range of values is given rather than a single data point. First, find the total frequency.
n = 339 + 86 + 55 + 18 + 11 + 8 + 3
n = 520
The mean is the expected value using the midpoints of each range.
μ = (339×100 + 86×300 + 55×500 + 18×700 + 11×900 + 8×1100 + 3×1300) / 520
μ = 122400 / 520
μ = 235.38
The variance is:
σ² = [(339×100² + 86×300² + 55×500² + 18×700² + 11×900² + 8×1100² + 3×1300²) − (520×235.38²)] / (520 − 1)
σ² = 55009.7
The standard deviation is:
σ = 234.54
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Answer:
Option B:
Step-by-step explanation:
The parabola has its concavity downwards, so we need a function in the model:
With a negative value of 'a'
The vertex is (0,0), so we have that:
The x-coordinate of the vertex is given by the equation:
So we have a function in the model:
With a < 0
The only option with this format is B:
