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DaniilM [7]
3 years ago
14

The average of five distinct scores has the same value as the median of the five scores. The sum of the five scores is 420. What

is the sum of the five scores that is not the median.
Mathematics
1 answer:
givi [52]3 years ago
3 0

Answer:

The sum of five scores that is not the median is, 336

Step-by-step explanation:

Average defined as the  the sum of the observation to the total number of the observation.

Given: The sum of the five scores is 420

Then, by definition of average;

Average of five distinct score = \frac{420}{5} = 84

It is also, given that the average of five distinct scores has the same value as the median of the five scores.

⇒Median of five scores = 84

Since, scores are distinct, therefore

Sum of all scores that is not the median = Sum of five scores - median = 420-84 = 336

therefore, the sum of five scores that is not the median is, 336

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A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.
nika2105 [10]

Answer:

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

\bar X=2.55 represent the sample mean for the late time for a flight

\mu population mean

\sigma=12 represent the population deviation

n=76 represent the sample size  

Confidence interval

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

The confidence interval for the true mean is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The Confidence level given is 0.95 or 95%, th significance would be \alpha=0.05 and \alpha/2 =0.025. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got z_{\alpha/2}=1.96

Replacing we got:

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

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3 years ago
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