Question

during one shift, the express lane clerk recorded how many times customers violated the \10 items or less\ rule for his lane. in particular, he recorded how many items over the limit each violator placed on the conveyor belt. this data is summarized in the histogram below. note: the last class actually represents \7 or more items,\ not just 7 items. what is the frequency of times the limit was exceeded by at least 2 items? ans =

Explanation:

Step1: Identify relevant classes

We need to find the classes where the number of items over 10 is at least 2. Looking at the x - axis labels (0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5), the class boundaries:
- The class for "at least 2 items over" starts from the class with lower boundary 2.5 (since 0.5 - 1.5 is 0 - 1 items over, 1.5 - 2.5 is 1 - 2 items over? Wait, actually, to be at least 2 items over, we need the number of items over 10 to be ≥2. So we need to consider the classes where the mid - point or the range is for 2 or more items over. Looking at the histogram, the classes are:
- 0.5 - 1.5: items over 10 is 0 - 1 (since the lower limit is 0.5, so 0.5 - 1.5 would correspond to 1 - 2? Wait, maybe better to look at the frequency for each class. Let's list the frequencies:
From the histogram:
- Class 0.5 - 1.5: frequency = 6 (items over 10: 1 - 2? No, maybe 0 - 1? Wait, the x - axis is "# items over 10". So 0.5 - 1.5: items over 10 is between 0.5 and 1.5, so approximately 1 item over? Wait, no, maybe the classes are:
- 0.5 - 1.5: 1 item over (since 0.5 to 1.5, so the number of items over 10 is 1 (because if you have 1 item over, it's between 0.5 and 1.5? Maybe the classes are:
- 0.5 - 1.5: 1 item over (frequency 6)
- 1.5 - 2.5: 2 items over (frequency 8)
- 2.5 - 3.5: 3 items over (frequency 13)
- 3.5 - 4.5: 4 items over (frequency 12)
- 4.5 - 5.5: 5 items over (frequency 12)
- 5.5 - 6.5: 6 items over (frequency 6)
- 6.5 - 7.5: 7 or more items over (frequency 5)

Wait, to be "at least 2 items over", we need the number of items over 10 to be ≥2. So we exclude the classes where items over 10 is less than 2 (i.e., 0 or 1 item over). The classes with items over 10 ≥2 are:

- 1.5 - 2.5: items over 10 = 2 (frequency 8)
- 2.5 - 3.5: items over 10 = 3 (frequency 13)
- 3.5 - 4.5: items over 10 = 4 (frequency 12)
- 4.5 - 5.5: items over 10 = 5 (frequency 12)
- 5.5 - 6.5: items over 10 = 6 (frequency 6)
- 6.5 - 7.5: items over 10 = 7 or more (frequency 5)

Wait, no. Wait, the first class (0.5 - 1.5) is for 1 item over (since 0.5 to 1.5, so the number of items over 10 is 1). The second class (1.5 - 2.5) is for 2 items over. So to be at least 2 items over, we need to include the classes where the number of items over 10 is 2 or more. So we need to sum the frequencies of the classes where items over 10 ≥2.

So the classes are:

- 1.5 - 2.5: frequency = 8 (2 items over)
- 2.5 - 3.5: frequency = 13 (3 items over)
- 3.5 - 4.5: frequency = 12 (4 items over)
- 4.5 - 5.5: frequency = 12 (5 items over)
- 5.5 - 6.5: frequency = 6 (6 items over)
- 6.5 - 7.5: frequency = 5 (7 or more items over)

Now, sum these frequencies: 8 + 13+12 + 12+6 + 5.

Step2: Sum the frequencies

First, 8+13 = 21; 21 + 12 = 33; 33+12 = 45; 45 + 6 = 51; 51+5 = 56.

Wait, but wait, maybe I made a mistake in the first class. Let's re - examine:

The x - axis is "# items over 10". So:

- 0.5 - 1.5: items over 10 is 1 (since 0.5 to 1.5, so the number of items over 10 is 1)
- 1.5 - 2.5: items over 10 is 2
- 2.5 - 3.5: items over 10 is 3
- 3.5 - 4.5: items over 10 is 4
- 4.5 - 5.5: items over 10 is 5
- 5.5 - 6.5: items over 10 is 6
- 6.5 - 7.5: items over 10 is 7 or more

So "at least 2 items over" means items over 10 ≥2. So we need to exclude the class where items over 10 is 1 (the 0.5 - 1.5 class with frequency 6) and the class where items over 10 is 0? No, the first class (0.5 - 1.5) is items over 10 = 1 (frequency 6), the second class (1.5 - 2.5) is items over 10 = 2 (frequency 8). Wait, if we want at least 2 items over, that means items over 10 ≥2. So the number of items over 10 can be 2,3,4,5,6,7 or more. So we need to sum the frequencies of the classes where items over 10 is 2 or more.

So the classes are:

- 1.5 - 2.5: frequency = 8 (items over 10 = 2)
- 2.5 - 3.5: frequency = 13 (items over 10 = 3)
- 3.5 - 4.5: frequency = 12 (items over 10 = 4)
- 4.5 - 5.5: frequency = 12 (items over 10 = 5)
- 5.5 - 6.5: frequency = 6 (items over 10 = 6)
- 6.5 - 7.5: frequency = 5 (items over 10 = 7 or more)

Now sum these frequencies: 8+13 + 12+12+6 + 5.

8+13 = 21; 21+12 = 33; 33+12 = 45; 45+6 = 51; 51+5 = 56.

Wait, but let's check again. Maybe the first class (0.5 - 1.5) is 0 - 1 items over (frequency 6), 1.5 - 2.5 is 1 - 2 items over (frequency 8), no, that can't be. Wait, the x - axis is "# items over 10", so the value on the x - axis is the number of items over 10. So a value of 0.5 on the x - axis means 0.5 items over 10, which doesn't make sense. So probably, the classes are:

- 0.5 - 1.5: 1 item over (since you can't have half an item, so it's 1 item over, frequency 6)
- 1.5 - 2.5: 2 items over (frequency 8)
- 2.5 - 3.5: 3 items over (frequency 13)
- 3.5 - 4.5: 4 items over (frequency 12)
- 4.5 - 5.5: 5 items over (frequency 12)
- 5.5 - 6.5: 6 items over (frequency 6)
- 6.5 - 7.5: 7 or more items over (frequency 5)

So to find the frequency of at least 2 items over, we need to sum the frequencies of all classes where the number of items over 10 is ≥2. So that's the sum of frequencies from the class with 2 items over (1.5 - 2.5) up to 7 or more items over (6.5 - 7.5).

So sum = 8 (2 items over) + 13 (3 items over) + 12 (4 items over) + 12 (5 items over) + 6 (6 items over) + 5 (7 or more items over).

Calculating: 8 + 13 = 21; 21+12 = 33; 33+12 = 45; 45+6 = 51; 51+5 = 56.

Wait, but let's check the total number of data points. The total frequency is 6 (0.5 - 1.5) + 8 (1.5 - 2.5)+13 + 12+12+6 + 5=6+8 = 14; 14+13 = 27; 27+12 = 39; 39+12 = 51; 51+6 = 57; 57+5 = 62. Wait, maybe I misread the frequencies. Let's look at the histogram again:

The y - axis (frequency) has 2,4,6,8,10,12,14.

- The first bar (0.5 - 1.5) reaches up to 6.
- The second bar (1.5 - 2.5) reaches up to 8.
- The third bar (2.5 - 3.5) reaches up to 13 (since 12 is a mark, and it's above 12, maybe 13).
- The fourth bar (3.5 - 4.5) reaches up to 12.
- The fifth bar (4.5 - 5.5) reaches up to 12.
- The sixth bar (5.5 - 6.5) reaches up to 6.
- The seventh bar (6.5 - 7.5) reaches up to 5.

So to find "at least 2 items over", we need to exclude the first two classes? No, wait:

If "at least 2 items over" means that the number of items over 10 is 2 or more. So the number of items over 10 can be 2,3,4,5,6,7 or more. So we need to sum the frequencies of the classes where the number of items over 10 is 2 or more.

The class for 2 items over is 1.5 - 2.5 (frequency 8), 3 items over is 2.5 - 3.5 (frequency 13), 4 items over is 3.5 - 4.5 (frequency 12), 5 items over is 4.5 - 5.5 (frequency 12), 6 items over is 5.5 - 6.5 (frequency 6), 7 or more items over is 6.5 - 7.5 (frequency 5).

So sum these frequencies: 8 + 13+12 + 12+6 + 5.

8+13 = 21; 21+12 = 33; 33+12 = 45; 45+6 = 51; 51+5 = 56.

Answer:

56