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Tems11 [23]
3 years ago
5

In performing a chi-square goodness-of-fit test for a normal distribution, a researcher wants to make sure that all of the expec

ted cell frequencies are at least five. The sample is divided into 7 intervals. The second through the sixth intervals all have expected cell frequencies of at least five. The first and the last intervals have expected cell frequencies of 1.5 each. After adjusting the number of intervals, the degrees of freedom for the chi-square statistic is____________.a. 2b. 3c. 5d. 7
Mathematics
1 answer:
Sholpan [36]3 years ago
3 0

Answer:

The degrees of freedom for the chi square statistic is 7-4=3

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Answer:

500

Step-by-step explanation:

Given: P = 150 - T where P is passenger cars and T is trucks

10P = 7T ⇔ P = \frac{7T}{10}

Plug in equation 2 into equation 1:

\frac{7T}{10} = 150 - T

Solving for T, we get 500

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Step-by-step explanation:

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12.4. Measure of Dispersion<br>2076 Q.No. 15 Find the standard deviation of: 4, 6, 8, 10, 12.​
Alenkinab [10]

Answer:

Standard deviation of given data = 3.16227

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Given sample size 'n' = 5

Given data  4, 6,8,10,12

Mean = \frac{4+6+8+10+12}{5} = 8

Mean of the sample x⁻ = 8

Standard deviation of the sample

                  S.D = \sqrt{\frac{Sum(x-x^{-} )^{2} }{n-1}}

<u><em>Step(ii)</em></u>:-

Given data

x          :         4      6       8       10      12

x-x⁻      :      4 - 8   6-8   8-8    10-8    12-8

(x-x⁻)   :        -4      -2     0          2        4

(x-x⁻)²  :        16     4       0         4        16  

 

  S.D = \sqrt{\frac{Sum(x-x^{-} )^{2} }{n-1}}

  S.D = \sqrt{\frac{16+4+0+4+16}{4}}

 S.D = √10 = 3.16227

<u><em> Final answer</em></u>:-

The standard deviation = 3.16227

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3 years ago
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Which of the following triangle cases may have one, two, or zero solutions?
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