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grin007 [14]
1 year ago
8

Let x be a random variable representing the amount of sleep each adult in New York City got last night. Consider a sampling dist

ribution of sample means x. (a) As the sample size becomes increasingly large, what distribution does the x distribution approach? uniform distribution sampling distribution normal distribution binomial distribution (b) As the sample size becomes increasingly large, what value will the mean μx of the x distribution approach? μx μ μ/√n μ/n σ (c) What value will the standard deviation σx of the sampling distribution approach? σ/n σx σ/√n μ σ (d) How do the two x distributions for sample size n = 50 and n = 100 compare? (Select all that apply.) The standard deviations are μ / 50 and μ / 100, respectively. The means are the same. The standard deviations are the same. The standard deviations are σ / √50 and σ / √100, respectively. The standard deviations are μ / √50 and μ / √100, respectively. The standard deviations are σ / 50 and σ / 100, respectively.
Mathematics
2 answers:
Sonbull [250]1 year ago
8 0

Answer:

a) As the sample size becomes increasingly large, the distribution approximates a normal distribution.

b) As the sample size becomes increasingly large, the mean of the sampling distribution, μₓ, approach as the population mean, μ.

σₓ

c) The standard deviation of the sampling distribution is given as

σₓ = (σ/√n)

d) Two sampling distributions with Sample size 50 and 100 have standard deviation of sampling distribution of (σ/√50) and (σ/√100) respectively.

Step-by-step explanation:

The Central limit theorem explains that sample distributions obtained from a population distribution (especially normally distributed ones) have distributions that approximate a normal distribution, a mean that is equal to the population mean and a standard deviation of sampling distribution given as the population standard deviation divided by square root of sample size.

a) Just like the Central limit theorem explains, as the sample size becomes increasingly large, the distribution approximates a normal distribution

b) And as the sample size becomes increasingly large, the mean of the sampling distribution, μₓ, approach as the population mean, μ.

c) The standard deviation of the sampling distribution is given as the population standard deviation divided by square root of sample size.

d) Using the formula given in (c) above, the standard deviation of sampling distribution of two sampling distributions with sample size 50 and 100 are (σ/√50) and (σ/√100) respectively.

Hope this Helps!!!

dangina [55]1 year ago
4 0

Answer:

Step-by-step explanation:

Given that X is a random variable representing the amount of sleep each adult in New York City got last night.

a) Normal distribution (by central limit theorem)

b) Sample mean will approach population mean mu

c) the standard deviation σx of the sampling distribution approach to σ/√n

d)  The means are the same.

. The standard deviations are σ / √50 and σ / √100, respectively.

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39,000/1,300 = 30. 30 x 4 = 120 Hours
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Suppose that 20% of the adult women in the United States dye or highlight their hair. We would like to know the probability that
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Answer:

71.08% probability that pˆ takes a value between 0.17 and 0.23.

Step-by-step explanation:

We use the binomial approxiation to the normal 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)}

Normal probability distribution

Problems of normally distributed samples 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)}.

In this problem, we have that:

p = 0.2, n = 200. So

\mu = E(X) = np = 200*0.2 = 40

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.2*0.8} = 5.66

In other words, find probability that pˆ takes a value between 0.17 and 0.23.

This probability is the pvalue of Z when X = 200*0.23 = 46 subtracted by the pvalue of Z when X = 200*0.17 = 34. So

X = 46

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

Z = \frac{46 - 40}{5.66}

Z = 1.06

Z = 1.06 has a pvalue of 0.8554

X = 34

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

Z = \frac{34 - 40}{5.66}

Z = -1.06

Z = -1.06 has a pvalue of 0.1446

0.8554 - 0.1446 = 0.7108

71.08% probability that pˆ takes a value between 0.17 and 0.23.

6 0
1 year ago
The Boeing 757-200 ER airliner carries 200 passengers and has doors with a height of 72 inches. Heights of men are normally dist
Maksim231197 [3]

Answer:b)0.8577

Step-by-step explanation:

Since the heights of men are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = heights of men

u = mean height

s = standard deviation

From the information given,

u = 69 inches

s = 2.8 inches

We want to find the probability that the mean height of the 100 men is less than 72 inches.. It is expressed as

P(x < 72)

For x = 72

z = (72 - 69)/2.8 = 1.07

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P(x < 72) = 0.8577

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Problem
Fiesta28 [93]

Answer:

do you have a photo of the figure?

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liberstina [14]

Answer: A. A(1) = 14; A(n) = (n − 1) −4; A(n) = 14 + (n − 1)(−4)

Step-by-step explanation:

Arithmetic sequence is a sequence that is identified by their common difference. Let a be the first term, n be the number of terms and d be the common difference.

For an arithmetic sequence, common difference 'd' is added to the preceding term to get its succeeding term. For example if a is the first term of a sequence, second term will be a+d, third term will give a+d+d and so on to generate sequence of the form,

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Notice that each new term keep increasing by a common difference 'd'

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Tn = 14+(n-1)-4 (which gives the required answer)

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