A biologist was keeping careful records of the time required, measured to the ne
ID: 3318273 • Letter: A
Question
A biologist was keeping careful records of the time required, measured to the nearest minute, for a spore culture to double in a petri dish. The number of minutes required by a sample of 14 determinations is provided below: 52 48 64 76 49 68- 67 60 70 7559 (4 marks) Construct a stem-and-leaf display for the data. Stem Leat a. b. Does the stem-and leaf display indicate that the data is skewed? If so, in what direction (2 mark (2 mark Calculate the mean time required for a spore culture to double. c. (2 ma d. Calculate the median time required for a spore culture to double math 215 imerev9.docExplanation / Answer
Part a
Required stem and leaf display for the data is given as below:
Stem
Leaves
4
8, 9
5
2, 9
6
0, 4, 7, 8, 9
7
0, 2, 5, 6, 7
Part b
Yes, stem and leaf display indicates that the data is skewed. It is skewed towards left side or it is negatively skewed.
Part c
From given data, we have
Total sum = 906
Total number of observations = 14
Mean = total sum / total number of observations = 906/14 = 64.71428571
Mean time required for a spore culture to double is 64.71 minutes.
Part d
From the given data,
Median = middle most observation when n is odd
Median = average of middle two observations when n is even
We are given n = 14, so n is even.
Median time = average of 7th and 8th observation when data is in increasing order.
Observations in an increasing order are given as below:
No.
X
1
48
2
49
3
52
4
59
5
60
6
64
7
67
8
68
9
69
10
70
11
72
12
75
13
76
14
77
7th observation = 67
8th observation = 68
Median = (67 + 68) / 2 = 67.5
Median time required for a spore culture to double = 67.5 minutes
Part e
Mode is the observation with maximum frequencies. For the given data, no any single observation is repeated in the data. So, modal time for a spore culture is absent.
Part f
Third quartile = average of 10th and 11th observation when data is in an increasing order.
We are given n = 14, for Q3, ¾ observations are below Q3 and ¼ observations are above Q3.
We have ¾ = 0.75 of 14 = 10.5, this means average of 10th and 11th observation.
10th observation = 70
11th observation = 72
Third quartile = Q3 = (70 + 72) / 2 = 71
Stem
Leaves
4
8, 9
5
2, 9
6
0, 4, 7, 8, 9
7
0, 2, 5, 6, 7
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