Question

1. Water samples from a certain river in North Carolina were taken at 16 different locations and tested for lead. Here are the data:

sample	lead (mg/L)
1	0.23
2	0.39
3	0.17
4	0.20
5	0.18
6	0.16
7	0.25
8	0.14
9	0.00
10	0.17
11	0.00
12	0.06
13	0.10
14	0.02
15	0.09
16	0.12

1. If the lead contamination level goal is 0.00 mg/L, what percentage of these river locations need decontamination work?

2. Use technology to make a histogram of the lead data. Use bins of width 0.05 starting at 0.00. Either paste your histogram into your work or sketch it carefully (whichever works better for you.)

3. Is the distribution unimodal, bimodal, multimodal, or uniform?

4. Is the distribution symmetric, skewed right, or skewed left?

Answer 1

Lead contamination level goal is 0.00 mg/L so river with lead contamination level higher than 0.00 mg/L need decontamination work.

Percentage of river locations that need decontamination work is (number of river location that need decontamination work/total number of river locations) × 100.

Observe the above data. Out of 16 river locations only 2 river locations have 0.00 mg/L contamination level so remaining 14 river locations need decontamination work so percentage of river locations that need decontamination work is (14/16) × 100 = 87.5.

So, 87.5% of these river locations need decontamination work.

We will use R software to obtain a histogram for the data.

R software codes

>x=c(0.23,0.39,0.17,0.20,0.18,0.16,0.25,0.14,0.00,0.17,0.00,0.06,0.10,0.02,0.09,0.12)

>hist(x, breaks=seq(0.00,0.4,0.05),col="pink")

Data vector x is created. hist() function is used to obtain histogram for the data. In the hist() function breaks is used to set bins width as 0.05.

R software output

Number of peaks in the histogram is the number of modes. If there is only one peak then the distribution is unimodal, if number of peaks is two then the distribution is bimodal and if the number of peaks is more than two then the distribution is multimodal.

Observe the above histogram, the peak is at one point only hence the distribution is unimodal.

A distribution is said to be right skewed if the right tail is longer and the distribution is said to be skewed left if the left tail is longer.

Since there are only 16 data points so we cannot precisely tell whether the distribution is symmetric, skewed left or skewed right. Observe the above histogram, it seems the distribution has long right tail hence we may say that the distribution is skewed right.

This answers your question. If you understood, please rate positively.

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