15.8k views
4 votes
The prices paid for a model of a new car are approximately normally distributed with a mean of $17,000 and a standard deviation of $500. The price that is 3 standard deviations above the mean is $ . The price that is 2 standard deviations below the mean is $ . The percentage of buyers who paid between $16,500 and $17,500 is % The percentage of buyers who paid between $17,000 and $18,000 is %. The percentage of buyers who paid less than $16,000 is %.

1 Answer

3 votes

Answer:

  • +3σ = $18,500
  • -2σ = $16,000
  • [-1σ, +1σ] = 68%
  • [0,+2σ] = 47.5%
  • (-∞, -2σ] = 2.5%

Explanation:

You want the percentages of buyers who paid prices in these intervals given the distribution of amounts paid is normal with mean $17000 and standard deviation $500:

  • [$16500, $17500]
  • [$17000, $18000]
  • [$0, $16000]

And you want the prices that are 3σ above and 2σ below the mean.

Empirical rule

The prices in this question all differ from the mean by integer multiples of the standard deviation. This suggests that the answers are supposed to be found using the empirical rule, which tells us the fraction of the distribution within ±1σ of the mean is 68%, and within ±2σ is 95%.

A) 3σ Above

The price at µ+3σ is $17000 +3(500) = $18,500.

B) 2σ Below

The price at µ-2σ is $17000 -2($500) = $16,000.

C) $16500–$17500

These prices are µ±1σ, so the empirical rule tells us 68% of buyers paid in this range.

D) $17000–$18000

This is half the number that lie between -2σ and +2σ, so the empirical rule tells us 95%/2 = 47.5% of buyers paid in this range.

E) 0–$16000

This is half the number that lie outside of ±2σ, so the empirical rule tells us (1 -95%)/2 = 2.5% of buyers paid in this range.

__

Additional comment

If you want more precise values, a probability calculator can provide them. The attached calculator display shows the percentages for C, D, E to be 68.3%, 47.7%, and 2.3%, respectively.

<95141404393>

The prices paid for a model of a new car are approximately normally distributed with-example-1
User Bas Jansen
by
8.4k points