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4 years ago

13

A set of six single-digit integers has a unique mode of 2. What is the largest possible

2 answers:

Sliva [168]4 years ago

6 0

Answer:

6.5

Step-by-step explanation:

We can assume that 2 is the smallest integer in the set. That means we would have a minimum of 2 2s. Then the largest integer in the set is 9. Since we can't have more than one 9 (due to the mode of 2), the next largest number we can have is 8. After that would be 7. Then the last digit would be 6. So now the set contains { 2, 2, 6, 7, 8, 9}. To find the median in an even set, we need to take the middle numbers and find the average. In this case, the two numbers are 6 and 7. The average of 6 and 7 is 6.5.

lbvjy [14]4 years ago

6 0

Answer:

the answer is 6.5

Step-by-step explanation:

your welcome :)

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3 years ago

The weights of 67 randomly selected axles were found to have a variance of 3.85. Construct the 80% confidence interval for the p

VLD [36.1K]

Answer:

$3.13$

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 67

Variance = 3.85

We have to find 80% confidence interval for the population variance of the weights.

Degree of freedom = 67 - 1 = 66

Level of significance = 0.2

Chi square critical value for lower tail =

$\chi^2_{1-\frac{\alpha}{2}}= 51.770$

Chi square critical value for upper tail =

$\chi^2_{\frac{\alpha}{2}}= 81.085$

80% confidence interval:

$\dfrac{(n-1)S^2}{\chi^2_{\frac{\alpha}{2}}} < \sigma^2 < \dfrac{(n-1)S^2}{\chi^2_{1-\frac{\alpha}{2}}}$

Putting values, we get,

$=\dfrac{(67-1)3.85}{81.085} < \sigma^2 < \dfrac{(67-1)3.85}{51.770}\\\\=3.13$

Thus, (3.13,4.91) is the required 80% confidence interval for the population variance of the weights.

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3 years ago