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Lunch break: In a recent survey of 655 working Americans ages 25-34, the average weekly amount spent on lunch was $43.5
with standard deviation $2.77. The weekly amounts are approximately bell-shaped.
Part 1 of 3
(a) Estimate the percentage of amounts that are between $35.26 and $51.88.
Almost all
of the amounts fall between $35.26 and $51.88.
Part 2 of 3
(b) Estimate the percentage of amounts that are between $38.03 and $49.11.
Approximately 95%
of the amounts fall between $38.03 and $49.11.
Part: 2/3
Part 3 of 3
(c) Between what two values will approximately 68% of the amounts be?
Approximately 68% of the amounts fall between
and S


Sagot :

Using the Empirical Rule, it is found that:

  • a) Approximately 99.7% of the amounts are between $35.26 and $51.88.
  • b) Approximately 95% of the amounts are between $38.03 and $49.11.
  • c) Approximately 68% of the amounts fall between $40.73 and $46.27.

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The Empirical Rule states that, in a bell-shaped distribution:

  • Approximately 68% of the measures are within 1 standard deviation of the mean.
  • Approximately 95% of the measures are within 2 standard deviations of the mean.
  • Approximately 99.7% of the measures are within 3 standard deviations of the mean.

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Item a:

[tex]43.5 - 3(2.77) = 35.26[/tex]

[tex]43.5 + 3(2.77) = 51.88[/tex]

Within 3 standard deviations of the mean, thus, approximately 99.7%.

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Item b:

[tex]43.5 - 2(2.77) = 38.03[/tex]

[tex]43.5 + 2(2.77) = 49.11[/tex]

Within 2 standard deviations of the mean, thus, approximately 95%.

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Item c:

  • 68% is within 1 standard deviation of the mean, so:

[tex]43.5 - 2.77 = 40.73[/tex]

[tex]43.5 + 2.77 = 46.27[/tex]

Approximately 68% of the amounts fall between $40.73 and $46.27.

A similar problem is given at https://brainly.com/question/15967965

It is often used for predicting results in statistics. This rule could be used as an estimate of the consequence of upcoming data to also be collected but also analyzed after determining a confidence interval and before accumulating precise data.

In a clock-shaped distribution, the empirical rule states:

  • About [tex]\bold{68\%}[/tex] of the measures are within 1 standard deviation of the mean.
  • About [tex]\bold{95\%}[/tex] of the measurements are within 2 standard deviations.
  • About [tex]\bold{99.7\%}[/tex] of the measures are subject to 3 standard deviations.

Point a:

[tex]\to \bold{43.5-3(2.77)= 43.5-8.31= 35.26}\\\\\to \bold{43.5+3(2.77)=43.5+8.31= 51.88}[/tex]

Therefore, approximately [tex]\bold{99.7\%}[/tex] within 3 standard deviations from the mean.

Point b:

[tex]\to \bold{43.5-2(2.77)= 43.5-5.54= 38.03}\\\\\to \bold{43.5+2(2.77)= 43.5+ 5.54= 49.11}\\\\[/tex]

Therefore, approximately [tex]\bold{95\%}[/tex] in 2 standard deviations from the mean.

Point c:

In 1 standard deviation of the mean, [tex]\bold{68\%}[/tex]is as follows:

[tex]\to \bold{43.5-2.77= 40.73}\\\\\to \bold{43.5+2.77= 46.27}[/tex]

About [tex]\bold{68\%}[/tex] of the sums decrease from [tex]\bold{\$40.73 \ and \ \$46.27}[/tex].

So, the final answer are:

  • The amount between [tex]\bold{\$35.26 \ and\ \$51.88}[/tex] makes up approximately [tex]\bold{99.7\%}[/tex] of the amounts.
  • The amount among [tex]\bold{\$ 38.03\ and\ \$49.11}[/tex] bills amounts to approximately [tex]\bold{95\%}.[/tex]
  • About [tex]\bold{68\%}[/tex] of the amounts are [tex]\bold{\$40.73 \ and \ \$46.27}[/tex].

Learn more:

brainly.com/question/14599976