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Maru [420]
2 years ago
6

What is 122.10 rounded to the nearest whole number?

Mathematics
1 answer:
I am Lyosha [343]2 years ago
6 0

Answer:

1

Round up if this number is greater than or equal to

5

and round down if it is less than

5

122

Hope I helped

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If you read can read 40 pages in 2 hours, how many pages can you read in 5 hours?
Pavlova-9 [17]
You can read 100 pages in 5 hours.

40 pages in 2 hours. We need to get the unit rate. In this case, we need to get the number of pages read in 1 hour.

40 pages ÷ 2 hours = 20 pages per hour.

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3 years ago
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Assume that 24.5% of people have sleepwalked. Assume that in a random sample of 1478 adults, 369 have sleepwalked. a. Assuming t
solniwko [45]

Answer:

a) 0.3483 = 34.83% probability that 369 or more of the 1478 adults have sleepwalked.

b) 369 < 403.4, which means that 369 is less than 2.5 standard deviations above the mean, and thus, a result of 369 is not significantly high.

c) Since the sample result is not significant, it suggests that the rate of 24.5% is a good estimate for the percentage of people that have sleepwalked.

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

A result is considered significantly high if it is more than 2.5 standard deviations above the mean.

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, 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)}.

Assume that 24.5% of people have sleepwalked.

This means that p = 0.245

Sample of 1478 adults:

This means that n = 1478

Mean and standard deviation:

\mu = 1478*0.245 = 362.11

\sigma = \sqrt{1478*0.245*0.755} = 16.5346

a. Assuming that the rate of 24.5% is correct, find the probability that 369 or more of the 1478 adults have sleepwalked.

Using continuity correction, this is P(X \geq 369 - 0.5) = P(X \geq 368.5), which is 1 subtracted by the p-value of Z when X = 368.5.

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

Z = \frac{368.5 - 362.11}{16.5346}

Z = 0.39

Z = 0.39 has a p-value of 0.6517

1 - 0.6517 = 0.3483

0.3483 = 34.83% probability that 369 or more of the 1478 adults have sleepwalked.

b. Is that result of 369 or more significantly high?

362.11 + 2.5*16.5346 = 403.4

369 < 403.4, which means that 369 is less than 2.5 standard deviations above the mean, and thus, a result of 369 is not significantly high.

c. What does the result suggest about the rate of 24.5%?

Since the sample result is not significant, it suggests that the rate of 24.5% is a good estimate for the percentage of people that have sleepwalked.

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