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Ber [7]
2 years ago
15

Some students reported how many pets they had. The dot plot shows the data collected: A dot plot is titled Students Pets and lab

eled Number of Pets. The values on the horizontal line are 0, 1, 2, 3, and 4. There are 5 dots above 0, 4 dots above 1, 3 dots above 2, 2 dot above 3, and 1 dot above 4. What do the dots above the number 3 indicate? There are exactly 2 students with 3 pets. There are exactly 3 students with 2 pets. There are exactly 2 students with more than 3 pets. There are exactly 3 students with more than 2 pets.
Mathematics
1 answer:
Kitty [74]2 years ago
8 0

Answer:

A dot plot is similar to histogram wherein they show the frequency of data. In this kind of plot, instead of bars, dots are stacked on their corresponding bin. The bins here are the number of pets, while the dots are the students. So, the 2 dots above number 3 signifies that, there are two students with three pets. The answer is A.

Step-by-step explanation:

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Step-by-step explanation:

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Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim
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The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}  

where x_i are data points, \bar{x} is the mean and n is the number of observations.  

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{10959}{25} =438.36

Sum of squares of differences = 175413.76

S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49

Population mean, μ = 425 mg/L

Sample mean, \bar{x} = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}

We use one-tailed t test to perform this hypothesis.

Formula:

t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }

Putting all the values, we have

t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813

Now, t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108

Since,                        

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

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