The following data gives the lengths, in cm, of 30 pieces of iron rods : 45 55 65 60 61 68 59 54 64 76 50 68 72 68 80 67 70 62 79 67 64 63 71 59 64 53 57 74...
Question 1 Report
The following data gives the lengths, in cm, of 30 pieces of iron rods :
45 55 65 60 61 68 59 54 64 76 50 68 72 68 80 67 70 62 79 67 64 63 71 59 64 53 57 74 55 57
(a) Using class intervals of 45 - 49, 50 - 54, 55 - 59, ... construct a frequency table of the data.
(b) Draw the histogram for the distribution
(c) Calculate the mean of the distribution
(d) What is the probability of selecting an iron rod whose length is in the modal class?
Answer Details
(a) Frequency table. Sorting the 30 lengths into the given class intervals and tallying gives:
| Length (cm) | Tally | Frequency \(f\) | Midpoint \(x\) | \(fx\) |
|---|---|---|---|---|
| 45 - 49 | | | 1 | 47 | 47 |
| 50 - 54 | ||| | 3 | 52 | 156 |
| 55 - 59 | |||| | | 6 | 57 | 342 |
| 60 - 64 | |||| || | 7 | 62 | 434 |
| 65 - 69 | |||| | | 6 | 67 | 402 |
| 70 - 74 | |||| | 4 | 72 | 288 |
| 75 - 79 | || | 2 | 77 | 154 |
| 80 - 84 | | | 1 | 82 | 82 |
| Total | 30 | 1905 |
(b) Histogram of the distribution. The bars are drawn against the class boundaries \(44.5, 49.5, 54.5, 59.5, 64.5, 69.5, 74.5, 79.5, 84.5\) on the horizontal axis, with heights equal to the class frequencies. Since all class widths are equal, the bar heights are simply the frequencies.
(c) Mean of the distribution.
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1905}{30}=63.5\text{ cm}.\](d) Probability of selecting a rod in the modal class. The modal class is the one with the highest frequency, \(60 - 64\) (frequency \(7\)). The number of rods in this class is \(7\) out of \(30\), so
\[P(\text{modal class})=\frac{7}{30}.\]Download The App On Google Playstore
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