For a given city, the graph represents the daily deviation, in °F, of the high temperature from the expected high...

Owning the Dataset

Table 1: Text Analysis

Table 2: Chart Analysis

Key Insights

• The majority of days (34 of 100) had high temperatures within \(\pm 2°\mathrm{F}\) of the expected seasonal value.
• Only 18 days (15 at \(+8°\mathrm{F}\), 3 at \(+12°\mathrm{F}\)) had high temperatures at least \(6°\mathrm{F}\) above expected (i.e., at least \(4°\mathrm{F}\) greater than seasonal).
• The temperature deviations are distributed symmetrically and in a bell-shaped pattern, with relatively few extreme outliers.
• All 100 days are accounted for by the data, confirming completeness of coverage.

Step-by-Step Solution

Question 1: Probability of High Temperature at Least 4°F Greater Than Seasonal

Complete Statement:

For a randomly selected day in this 100-day period, the probability that the high temperature was at least \(4°\mathrm{F}\) greater than seasonal is ________

Breaking Down the Statement

• Statement Breakdown 1:
  • Key Phrase: randomly selected day in this 100-day period
    • Meaning: We're asked for the probability over all 100 days shown in the histogram.
    • Relation to Chart: All bars in the histogram together represent the 100 total days.
    • Important Implications: Probability will be (number of favorable days) divided by 100.

• Statement Breakdown 2:
  • Key Phrase: at least \(4°\mathrm{F}\) greater than seasonal
    • Meaning: The temperature has to be at least \(4°\mathrm{F}\) above what counts as 'seasonal.'
    • Relation to Chart: According to the setup, 'greater than seasonal' means above the right endpoint (\(2°\mathrm{F}\)) of the class centered at 0. So 'at least \(4°\mathrm{F}\) greater than seasonal' means at least \((2+4) = 6°\mathrm{F}\) above expected.
    • Important Implications: The only days that count are those with temperature deviations of \(6°\mathrm{F}\) or more above expected.

• What is needed: The fraction of days in the histogram with deviations of at least \(6°\mathrm{F}\) above expected.

Solution:

• Condensed Solution Implementation:
  Find the histogram bars that represent deviations of \(6°\mathrm{F}\) or more (classes centered at \(8°\mathrm{F}\) and \(12°\mathrm{F}\)), sum their frequencies, and divide by 100.
• Necessary Data points:
  The bar for class centered at \(8°\mathrm{F}\) (\([6°\mathrm{F}, 10°\mathrm{F})\)) contains 15 days; the bar for class centered at \(12°\mathrm{F}\) (\([10°\mathrm{F}, 14°\mathrm{F})\)) contains 3 days.
  • Calculations Estimations:
    15 days (\(8°\mathrm{F}\) bin) + 3 days (\(12°\mathrm{F}\) bin) = 18 days; Probability: \(\frac{18}{100} = 0.18\).
  • Comparison to Answer Choices:
    The answer choices include 0.18. Our calculated probability matches this value.

FINAL ANSWER Blank 1: 0.18

Question 2: Probability of High Temperature at Least 8°F Greater Than Seasonal

Complete Statement:

and the probability that the high temperature was at least \(8°\mathrm{F}\) greater than the seasonal is _______

Breaking Down the Statement

• Statement Breakdown 1:
  • Key Phrase: at least \(8°\mathrm{F}\) greater than seasonal
    • Meaning: The temperature has to be at least \(8°\mathrm{F}\) above what counts as 'seasonal.'
    • Relation to Chart: Using the same logic as before, this means at least \((2+8) = 10°\mathrm{F}\) above expected.

• What is needed: The fraction of days in the histogram with deviations of at least \(10°\mathrm{F}\) above expected.

Solution:

• Condensed Solution Implementation:
  Find the histogram bar(s) for deviations of \(10°\mathrm{F}\) or more (class centered at \(12°\mathrm{F}\) only), sum their frequencies, and divide by 100.
• Necessary Data points:
  The bar for class centered at \(12°\mathrm{F}\) (\([10°\mathrm{F}, 14°\mathrm{F})\)) contains 3 days.
  • Calculations Estimations:
    3 days (\(12°\mathrm{F}\) bin); Probability: \(\frac{3}{100} = 0.03\).
  • Comparison to Answer Choices:
    The answer choices include 0.03, agreeing with our calculation.

FINAL ANSWER Blank 2: 0.03

Summary

By noting that '\(x°\mathrm{F}\) greater than seasonal' means \(x°\mathrm{F}\) above the right endpoint (\(2°\mathrm{F}\)) of the class centered at 0, we translate the worded description into a simple counting problem on the histogram. For at least \(4°\mathrm{F}\) above seasonal (i.e., \(\geq 6°\mathrm{F}\) above expected) there are 18 days (0.18 probability), and for at least \(8°\mathrm{F}\) above seasonal (i.e., \(\geq 10°\mathrm{F}\) above expected) there are 3 days (0.03 probability).

Question Independence Analysis

These questions are independent because each asks for the probability of a different temperature threshold above seasonal, and each is solved by counting the relevant days separately.