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guapka [62]
3 years ago
12

PLEASE HELP ME OUT ASAP! I WOULD REALLY APPRECIATE IT AND GOD BLESS

Mathematics
1 answer:
77julia77 [94]3 years ago
5 0
To approximate the P(x<27) we need to find the z-score of the data, this will be given by:
z=(x-μ)/σ
where:
μ-mean
σ-standard deviation
x=27, μ=32, σ=4
z=(27-32)/4
z=-5/4
z=-1.25
thus
P(x<27)=P(z<-1.25)
=0.1056
=10.56%

Answer: 10.56%
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Of the total population of American households, including older Americans and perhaps some not so old, 17.3% receive retirement
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Answer:

47.54% probability that more than 20 households but fewer than 35 households receive a retirement income

Step-by-step explanation:

We use the binomial aproxiation 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 = 0.173, n = 120. So

\mu = E(X) = np = 120*0.173 = 20.76

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.173*0.827} = 4.14

In a random sample of 120 households, what is the probability that more than 20 households but fewer than 35 households receive a retirement income?

We are working with discrete values, so this is the pvalue of Z when X = 35-1 = 34 subtracted by the pvalue of Z when X = 20 + 1 = 21.

X = 34

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

Z = \frac{34 - 20.76}{4.14}

Z = 3.2

Z = 3.2 has a pvalue of 0.9993

X = 21

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

Z = \frac{21 - 20.76}{4.14}

Z = 0.06

Z = 0.06 has a pvalue of 0.5239

0.9993 - 0.5239 = 0.4754

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