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Hitman42 [59]
3 years ago
8

A Food Marketing Institute found that 30% of households spend more than $125 a week on groceries. Assume the population proporti

on is 0.3 and a simple random sample of 245 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is less than 0.32
Mathematics
1 answer:
MaRussiya [10]3 years ago
6 0

Answer:

75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32

Step-by-step explanation:

Problems of normally distributed samples are 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.

For the sampling distribution of a proportion p with size n, we have that \mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}

In this problem, we have that:

p = 0.3, n= = 245

So

\mu = 0.3, \sigma = \sqrt{\frac{0.3*0.7}{245}} = 0.0293

What is the probability that the sample proportion of households spending more than $125 a week is less than 0.32

This is the pvalue of Z when X = 0.32. So

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

Z = \frac{0.32 - 0.3}{0.0293}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32

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Katen [24]

Answer:

what? i belive the answer is B

Step-by-step explanation:

4 0
2 years ago
The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit
Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

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

By the Central Limit Theorem

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

Z = \frac{160 - 159}{1.68}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257

X = 150

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

Z = \frac{158 - 159}{1.68}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

7 0
3 years ago
A company surveyed 2,600 North American airline passengers and reported that approximately 75% said that they carry a smartphone
geniusboy [140]

The probability that exactly six out of ten carry a smartphone when travelling is; <u><em>P(6) = 0.146</em></u>

The missing part of the question is;

The probability distribution of x is the binomial distribution with n = 10 and p = 0.75.

Calculate P(6)

  • This is a binomial <em>probability distribution</em> problem that has a general formula as; P(x) = ⁿCₓ × pˣ × q⁽ⁿ ⁻ ˣ⁾

Where;

p is probability of success

q is probability of failure

n is number of experiments

In this question;

n = 10

p = 75% = 0.75

q = 1 - p

q = 1 - 0.75

q = 0.25

Since six passengers are randomly selected, then applying the formula earlier quoted, we have;

P(6) = ¹⁰C₆ × 0.75⁶ × 0.25⁽¹⁰ ⁻ ⁶⁾

P(6) = 0.146

Read more at; brainly.com/question/24239758

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1/4 cup equals how many pounds?
Leno4ka [110]

1/4 cup equals 0.25 lbs.

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What are the values of each Halloween icon? (Math Logic Puzzles) (78 POINTS)
Mekhanik [1.2K]

Answer:

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