Question

ex5) make a histogram for the given data. then describe the shape of its distribution.
data table:
25, 43, 65, 12, 8
30, 44, 68, 18, 21
35, 33, 37, 54, 61
29, 31, 38, 22, 48
19, 34, 55, 14, 37
intervals and frequency table (empty) and a histogram drawing area (empty)

Explanation:

Step1: Find the range of data

First, identify the minimum and maximum values in the data set. The data points are: 25, 43, 65, 12, 8, 30, 44, 68, 18, 21, 35, 33, 37, 54, 61, 29, 31, 38, 22, 48, 19, 34, 55, 14, 37.
The minimum value ($\text{min}$) is 8, and the maximum value ($\text{max}$) is 68.
The range is $\text{max} - \text{min} = 68 - 8 = 60$.

Step2: Determine the number of intervals (classes)

A common rule is to use 5 - 10 classes. Let's choose 5 intervals for simplicity. The class width ($w$) is calculated as $\frac{\text{range}}{\text{number of classes}}=\frac{60}{5} = 12$. We can adjust slightly for convenience, so let's use a class width of 12 (or 10 - 15, but 12 works here).

Step3: Define the intervals

Start with the minimum value. The intervals will be:
- 8 - 19 (since 8 is the min, and 8 + 11 = 19, next interval starts at 20)
- 20 - 31 (20 to 31, width 12)
- 32 - 43 (32 to 43)
- 44 - 55 (44 to 55)
- 56 - 67 (56 to 67)
- 68 - 79 (68 is the max, so last interval: 68 - 79, but 68 is only one point here) Wait, maybe better to use 6 intervals with width 10:
- 8 - 17
- 18 - 27
- 28 - 37
- 38 - 47
- 48 - 57
- 58 - 67
- 68 - 77 (but 68 is one point). Alternatively, let's list all data points and count frequencies:

Data points:
8,12,65,25,43,30,44,68,18,21,35,33,37,54,61,29,31,38,22,48,19,34,55,14,37

Now, let's create intervals (let's use 5 intervals with width 12, adjusted):

Interval 1: 8 - 19 (values: 8,12,18,19,14) → frequency = 5
Interval 2: 20 - 31 (values:25,30,21,29,31,22) → frequency = 6
Interval 3: 32 - 43 (values:35,33,37,34,37,43) → frequency = 6
Interval 4: 44 - 55 (values:44,54,48,55) → frequency = 4
Interval 5: 56 - 67 (values:65,61) → frequency = 2
Interval 6: 68 - 79 (value:68) → frequency = 1

Wait, maybe better to count properly:

List all data points:
8,12,14,18,19,21,22,25,29,30,31,33,34,35,37,37,38,43,44,48,54,55,61,65,68

Now, group into intervals (let's use 6 intervals with width 10):

1. 8 - 17: 8,12,14 → frequency = 3
2. 18 - 27: 18,19,21,22,25 → frequency = 5
3. 28 - 37: 29,30,31,33,34,35,37,37 → frequency = 8
4. 38 - 47: 38,43,44 → frequency = 3 (wait, 38,43,44? 44 is 44, which is in next interval. Oops, 38 - 47: 38,43 (43 is 43, 44 is 44>47? No, 47 is end. So 38,43 → frequency 2? Wait, 44 is in 48 - 57? No, 44 - 53? Maybe my interval definition was wrong. Let's use the formula:

Number of classes ($k$) can be found using Sturges' formula: $k = 1 + 3.322\log(n)$, where $n$ is the number of data points. Here, $n = 25$ (let's count: 5 rows × 5 columns? Wait the table has 5 rows (25,43,65,12,8; 30,44,68,18,21; 35,33,37,54,61; 29,31,38,22,48; 19,34,55,14,37) → 5×5=25 data points.

Sturges' formula: $k = 1 + 3.322\log(25) \approx 1 + 3.322×1.39794 \approx 1 + 4.64 \approx 5.64$, so 6 classes.

Class width: $\frac{68 - 8}{6} \approx 10$, so class width 10.

Intervals:
- 8 - 17 (8 to 17, inclusive or exclusive? Let's use inclusive of lower, exclusive of upper: (8,17]
- (17,27]
- (27,37]
- (37,47]
- (47,57]
- (57,67]
- (67,77] (but 68 is in (67,77])

Now count frequencies:

1. 8 - 17: 8,12,14 → 3
2. 18 - 27: 18,19,21,22,25 → 5
3. 28 - 37: 29,30,31,33,34,35,37,37 → 8 (28 to 37: 29,30,31,33,34,35,37,37)
4. 38 - 47: 38,43,44 → 3 (38,43,44; 44 is 44 < 47? 47 is end, so 44 is in (37,47]? Wait, 38 - 47: 38,43,44, and 48 is in next. Wait 44 is 44, so yes, 38,43,44 → 3
5. 48 - 57: 48,54,55 → 3 (48,54,55)
6. 58 - 67: 61,65 → 2 (61,65)
7. 68 - 77: 68 → 1 (68)

Now, to make the histogram, we plot the intervals on the x - axis and frequencies on the y - axis. Each bar's height is the frequency of the interval.

Step4: Describe the shape

After creating the histogram, we look at the distribution. The frequencies are: 3,5,8,3,3,2,1. The peak is in the 28 - 37 interval (frequency 8), then it decreases on both sides. So the distribution is somewhat symmetric? Wait, no, the left side (lower intervals) has lower frequencies, and the right side (higher intervals) also has lower frequencies, but the peak is in the middle. Wait, maybe slightly skewed right? Wait, the tail on the right (higher values) has a few points (61,65,68), so the right tail is longer? Wait, the frequencies for lower intervals (8 - 17:3, 18 - 27:5) are increasing towards the peak (28 - 37:8), then decreasing (38 - 47:3, 48 - 57:3, 58 - 67:2, 68 - 77:1). So the distribution is approximately symmetric with a peak in the middle (28 - 37), or slightly skewed right because the right tail (58 - 77) has a longer tail (more spread out) than the left tail (8 - 27). Alternatively, it's a unimodal distribution (one peak) with the peak around 28 - 37, and the data is somewhat symmetric but with a slight right skew.

Answer:

To create the histogram:

1. Intervals & Frequencies (example with 7 intervals, width 10):

Intervals	Frequency
18–27	5
28–37	8
38–47	3
48–57	3
58–67	2
68–77	1

1. Histogram Construction: Plot intervals on the x - axis, frequencies on the y - axis. Draw bars with heights equal to their interval’s frequency.

1. Shape Description: The distribution is unimodal (one peak) at 28–37, with a slight right skew (longer tail on the right - hand side, as higher intervals have fewer data points) or approximately symmetric around the peak.