Mathematics, 15.07.2021 18:40 Browardlovesangel

A Food Marketing Institute found that 34% of households spend more than $125 a week on groceries. Assume the population proportion is 0.34 and a simple random sample of 124 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 0.31

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Answer from: asjessehills2000

0.2405 = 24.05% probability that the sample proportion of households spending more than $125 a week is less than 0.31.

Step-by-step explanation:

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

Normal Probability Distribution

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

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Assume the population proportion is 0.34 and a simple random sample of 124 households is selected from the population.

This means that p = 0.34, n = 124

Mean and standard deviation:

$\mu = p = 0.34$

$s = \sqrt{\frac{0.34*0.66}{124}} = 0.0425$

What is the probability that the sample proportion of households spending more than $125 a week is less than 0.31?

This is the p-value of Z when X = 0.31, so:

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

$Z = \frac{0.31 - 0.34}{0.0425}$

Z = -0.705

Z = -0.705 has a p-value of 0.2405.

0.2405 = 24.05% probability that the sample proportion of households spending more than $125 a week is less than 0.31.

Answer from: Quest

step-by-step explanation:

yuo ned the activity or vertical angle of the form of 1,7 but 2,2 has a c4,2 as it verticie so i not have a car

Answer from: Quest

2 & 7

step-by-step explanation:

he climbs 300 ft/hr at a constant rate. the peak is at 1,500 ft.

when going up, he reaches 600 ft in 2 hours. remember, that he is going in a constant rate. divide 600 with 300:

600/300 = 2

assuming that he takes the same route back (meaning that it's a 1500 ft route back, you can solve.

note also that you are solving when he gets back to 600 ft after he reaches the peak.

first, find the amount of hours it takes to get up the peak. divide the total amount (1500), with the ft/her (300) to get the amount of hours (x):

1500/300 = x

x = 5

it takes jamie 5 hours to get up. now, he is returning down, and reaches 600 (his current destination). remember that he climbs at a constant 300 ft/hr. divide 600 with 300:

600/300 = x

x = 2

now add the two numbers together.

x = 5 + 2 = 7

jamie takes 7 hours to go up the mountain and reach his current position when descending.

(total amount/constant rate) = (time).

remember that it is a constant rate, so in total, when he gets back down the mountain, it will take the same amount of time as when he goes up. the total amount for the round trip is: 5 + 5 = 10.