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3 years ago

A person is going to fly from New York to Rio de Janeiro. If he leaves at 1:00 p.m., Eastern Standard Time and the filght

Mathematics

1 answer:

vesna_86 [32]3 years ago

8 0

Answer:

8:30

Step-by-step explanation:

At 12:00 (noon) In New York, It is 2:00 pm in Rio De Janerio.

That if he left at 1 in NY it would've been 3 in Rio, 3+5.5=8.5

Which is 8:30 pm. Hope this helped!!

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2 years ago

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3 years ago

What number is between 3.16 and 3.17?

VladimirAG [237]

There are multiple answers to this question
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3 years ago

What should you do? Jason collects baseball cards. He has 33 cards. He wants to put 4 cards on each page of his book. How many p

Nastasia [14]

Answer:

Increase the quotient by one

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Since in the question it is mentioned that There is a total of 33 cards and wants to put 4 cards on each page

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3 years ago

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.

3 0

3 years ago