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ivanzaharov [21]
3 years ago
15

Number Frequency 1,400 3 1,450 7 1,500 7 1,550 5 1,600 4 1,650 2 1,700 1 What is the mode?

Mathematics
2 answers:
Iteru [2.4K]3 years ago
4 0

Answer: 1,450 and 1,500

Step-by-step explanation:

  • The mode of a data set refers to the most-common or most-frequently occurring value in a series of data. There can be more than one mode in a data set.

The given table is

Number Frequency

1,400 3

1,450 7

1,500 7

1,550 5

1,600 4

1,650 2

1,700 1

In the given table, the most frequent values are 1,450 and 1,500 with a frequency of 7.

hence, the mode = 1,450 and 1,500

poizon [28]3 years ago
3 0
Mode is the number that shows up the most

in this case, it is 71,500 which shows up twice, one more than any of the other numbers

hope this helps
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About 12.5% of restaurant bills are incorrect. If 200 bills are selected at ran- dom, find the probability that at least 22 will
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Answer:

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Step-by-step explanation:

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Normal probability distribution

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In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

About 12.5% of restaurant bills are incorrect.

This means that p = 0.125

200 bills are selected at random

This means that n = 200

Mean and standard deviation:

\mu = E(X) = np = 200*0.125 = 25

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.125*0.875} = 4.677

Find the probability that at least 22 will contain an error.

Using continuity correction, this is P(X \geq 22 - 0.5) = P(X \geq 21.5), which is 1 subtracted by the p-value of Z when X = 21.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{21.5 - 25}{4.677}

Z = -0.75

Z = -0.75 has a p-value of 0.2266.

1 - 0.2266 = 0.7734

0.7734 = 77.34% probability that at least 22 will contain an error. Probability above 50%, which means that this is likely to occur.

8 0
3 years ago
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