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Finger [1]
3 years ago
10

Cindy struck out 7, 14, 6, 4, 5, and 6 batters in softball games this season. In discussing her stats with a newspaper reporter,

Cindy's coach mentions that the average number of Cindy's strikeouts is 7. Is the coach's statement misleading? Explain why or why not.
Mathematics
1 answer:
Murljashka [212]3 years ago
8 0

Answer:

The coach's statement is not misleading because her average strikeouts  is 7.

Where as the coach also tells that it is 7.  

Step-by-step explanation:

The average is taken out by adding the observations and dividing by the total number of observations.

X`= ∑x/n     where n is the number of observations.

Average = 7 + 14 + 6 + 4 +5 +6 /6

Average = 42/6= 7

The coach's statement is not misleading because her average strikeouts  is 7.

Where as the coach also tells that it is 7.  

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How do u turn 2 3/4 into a improper faction
n200080 [17]

To turn a mixed number into an improper fraction, you have to multiply the denominator with the whole number and add the numerator to the result. The denominator doesn't change.

The denominator is 4. The whole number is 2. Multiply:

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The numerator is 3. Add that to the product of the denominator and whole number.

8+3=11

The denominator stays the same. The denominator is 4. Make it as a fraction(improper).

Numerator = 11 \\ Denominator = 4 \\ \\ Fraction: \frac{11}{4}

Your answer is \frac{11}{4}

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A chess board is an 8 by 8 grid of squares. If you randomly choose 3 squares on the board, what is the probability that:
bixtya [17]
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The Ohio Department of Agriculture tested 203 fuel samples across the state
Rus_ich [418]

Answer:

\hat p = \frac{14}{105}= 0.133

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475

0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58

The proportion estimated would be:

\hat p = \frac{14}{105}= 0.133

And that represent the proportion of failures.

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475

0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185

The 99% confidence interval would be given by (0.0475;0.2185)

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aivan3 [116]
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1 forward from k is l

the next letter is L
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