The Whitcomb Company manufactures a metal ring for industrial engines that usually weighs about 50 ounces. A random sample of 50
of these metal rings produced the following weights (in ounces). 51 53 56 50 44 47 53 53 42 57 46 55 41 44 52 56 50 57 44 46 41 52 69 53 57 51 54 63 42 47 47 52 53 46 36 58 51 38 49 50 62 39 44 55 43 52 43 42 57 49
(a) Construct the frequency table and histogram for these data using eight classes. (You can choose your own classes)
(b) What can you observe about the data from the histogram? tssabout 500 nces. Doyoub elieve their claim? Explain your reason.
(a) Construct the frequency table and histogram for these data using eight classes. (You can choose your own classes)
(b) What can you observe about the data from the histogram? tssabout 500 nces. Doyoub elieve their claim? Explain your reason.
2 answers:
It’s important to begin by creating a frequency table to simplify plotting, subsequently organizing your data into classes for ease of visualization, such as 35-40, 41-45, etc.
B) Observations reveal a significant demand for metal rings weighing between 50-55 ounces and from 55-60 ounces.
Answers:
a) Please check the images below.
b) No.
Step-by-step explanation:
Hi,
A histogram resembles a bar chart but bins numbers into intervals.
The weights data signifies:
Range = Maximum Value - Minimum Value
= 69 - 36
= 33
Required classes are 8.
As we construct our data with this width, it will leave a few values out; hence, we slightly adjust the max and min values to ensure all values fit into the range.
Approximate max = 71
Approximate min = 35
New range = 71 - 35
= 36
Based on this, we formulate our frequency table:
Starting from 35, we add 4.5 to reach 39.5.
The next value is obtained by adding another 4.5 to 39.5, making it 44.0.
This continues until all classes are established:
35.0 - 39.5
39.5 - 44.0
44.0 - 48.5
48.5 - 53.0
53.0 - 57.5
57.5 - 62.0
62.0 - 66.5
66.5 - 71.0
With each class width, count how many values fall within those intervals to establish the frequency count.
The final table appears as displayed in the accompanying image below.
To build a histogram, plot the class intervals (x-axis) against the frequency (y-axis). The histogram will resemble the image below.
b)
A close inspection of the table indicates that frequencies are skewed more towards the left. Thus, we characterize this distribution as left-skewed.
Additionally, it can be inferred that the mean and other central values are pushed more towards the left side of the graph.
Hence, the claim that their metal rings weigh around 50 ounces seems unfounded as central values are predominantly left-skewed.
You might be interested in
The likelihood of selecting one girl is calculated as . This is based on having 5 girls within a total of 12 students, and the probability of an event can be expressed as:
.
Using the same reasoning, for the next student, we have reduced the number of students by 1, leading to 11 possible outcomes instead of 12, giving us:, which represents the probability of selecting a boy as the second choice.
Lastly, the probability of choosing a girl for the third selection follows the same logic and is given as:.
However, we must combine these individual probabilities to determine the likelihood of this specific sequence of selections occurring:
This simplifies to: