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Difficulty: MediumFrequency Distributions, Tables, and Grouped Data

The frequency distribution table below summarizes the monthly water consumption, cc (in cubic meters, m3\text{m}^3), recorded for a sample of 150150 municipal water accounts.

Monthly Water Consumption (m3\text{m}^3)Number of Accounts
0≤c<100 \le c < 102525
10≤c<2010 \le c < 204545
20≤c<3020 \le c < 305050
30≤c<4030 \le c < 402020
40≤c<5040 \le c < 501010

If one water account is selected at random from among all accounts with a monthly water consumption of at least 10 m310\text{ m}^3, what is the probability that the selected account has a monthly water consumption of less than 30 m330\text{ m}^3? (Give your answer as a decimal rounded to two decimal places.)

Answer: 0.76

Answer

0.76
To calculate the required probability, first restrict the sample space to accounts with a monthly consumption of at least 10 m310\text{ m}^3. Summing the frequencies for the intervals 10≤c<2010 \le c < 20, 20≤c<3020 \le c < 30, 30≤c<4030 \le c < 40, and 40≤c<5040 \le c < 50 gives 45+50+20+10=12545 + 50 + 20 + 10 = 125 accounts. Among these 125125 accounts, those with a consumption of less than 30 m330\text{ m}^3 fall into the intervals 10≤c<2010 \le c < 20 and 20≤c<3020 \le c < 30, giving a count of 45+50=9545 + 50 = 95 accounts. Dividing the favorable outcomes by the total outcomes in the restricted sample space yields 95125=0.76\frac{95}{125} = 0.76.

Step-by-Step Solution

1
Determine the total number of accounts meeting the condition of having consumption of at least 10 m310\text{ m}^3.
Total eligible accounts = 45+50+20+10=12545 + 50 + 20 + 10 = 125.
Accounts with consumption of at least 10 m310\text{ m}^3 fall into the intervals 10≤c<2010 \le c < 20, 20≤c<3020 \le c < 30, 30≤c<4030 \le c < 40, and 40≤c<5040 \le c < 50.
2
Determine the number of accounts among the eligible set with consumption less than 30 m330\text{ m}^3.
Number of favorable accounts = 45+50=9545 + 50 = 95.
Within the eligible set, accounts with consumption less than 30 m330\text{ m}^3 fall into the intervals 10≤c<2010 \le c < 20 and 20≤c<3020 \le c < 30.
3
Calculate the conditional probability as a decimal.
95125=0.76\frac{95}{125} = 0.76
Dividing the favorable outcomes (9595) by the total possible outcomes in the restricted sample space (125125) yields 0.760.76.

Key Concept

Conditional probability and sample space restriction in grouped frequency tables
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Topics

Frequency Distributions, Tables, and Grouped Data

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