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2 years ago

6

In January of 2002(group 1),700 out of the 1700 spots were bare ground (no vegetation). Find the sample proportion of bare groun

d spots.

1 answer:

dlinn [17]2 years ago

8 0

Answer:

The sample proportion of bare ground spots is 0.4118

Step-by-step explanation:

The sample proportion of bareground spots is the number of bareground sports divided by the number of spots.

In this question

700 bareground spots

1700 spots

7/17 = 0.4118

The sample proportion of bare ground spots is 0.4118

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A local company makes snack-size bags of potato chips. Each day, the company produces batches of 400 snack-size bags using a pro

ioda

Answer:

Probability that a sample of 40 bags has an average weight of at least 2.02 ounces is 0.103.

Step-by-step explanation:

We are given that the company produces batches of 400 snack-size bags using a process designed to fill each bag with an average of 2 ounces of potato chips. Assume the amount placed in each of the 400 bags is normally distributed and has a standard deviation of 0.1 ounce.

Also, sample of 40 bags are selected.

<em>Let </em> $\bar X$ <em> = sample average weight</em>

The z-score probability distribution for sample mean is given by ;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean weight of potato chips = 2 ounces

$\sigma$ = standard deviation = 0.1 ounces

n = sample of bags = 40

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, the probability that a sample of 40 bags has an average weight of at least 2.02 ounces is given by = P( $\bar X$ $\geq$ 2.02 ounces)

P( $\bar X$ $\geq$ 2.02) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{2.02-2}{\frac{0.1}{\sqrt{40} } }$ ) = P(Z $\geq$ 1.265) = 1 - P(Z < 1.265)

= 1 - 0.89707 = 0.103

<em>The above probability is calculated using z table by looking at value of x = 1.265 which will lie between x = 1.26 and x = 1.27 in the z table which have an area of 0.89707.</em>

<em />

Therefore, probability that a sample of 40 bags has an average weight of at least 2.02 ounces is 0.103.

3 0

2 years ago

Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se

Ugo [173]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

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:

$n = 175$

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that $p = 0.467$ . So

$\mu = E(X) = np = 175*0.467 = 81.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6$

This probability, using continuity correction, is $P(X \geq 85 - 0.5) = P(X \geq 84.5)$ , which is 1 subtracted by the pvalue of Z when X = 84.5. So

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

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

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

$Z = \frac{89.5 - 81.725}{6.6}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 79.5

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

$Z = \frac{79.5 - 81.725}{6.6}$

$Z = -0.34$

$Z = -0.34$ has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that $p = 0.063$

$\mu = E(X) = np = 175*0.063 = 11.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141$

This probability, using continuity correction, is $P(X < 12 - 0.5) = P(X < 11.5)$ , which is the pvalue of Z when X = 11.5. So

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

$Z = \frac{11.5 - 11.025}{3.2141}$

$Z = 0.15$

$Z = 0.15$ has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that $p = 0.227$

$\mu = E(X) = np = 175*0.227 = 39.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

Using continuity correction, this probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

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

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

8 0

2 years ago

Jo is going on an 8-day activity holiday. Each day she can choose one of the water sports: kayaking or sailing, or land-based sp

Ronch [10]

Answer:

6 ways

Step-by-step explanation:

Given

Number of Sports: 3

Required

Determine the total number of schedules

Since, there are 3 sports and the schedule is in no particular order.

The number of schedules is calculated as thus:

$Number = n!\ ways$

Where:

$n = 3$

So, we have:

$Number = 3 * 2 * 1\ ways$

$Number = 6\ ways$

3 0

2 years ago

A political polling agency wants to take a random sample of registered voters and ask whether or not they will vote for a certai

bonufazy [111]

Answer:

C. Different sample proportions would result each time, but for either sample size, they would be centered (have their mean) at the true population proportion.

Step-by-step explanation:

From the given information;

A political polling agency wants to take a random sample of registered voters and ask whether or not they will vote for a certain candidate.

A random sample is usually an outcome of any experiment that cannot be predicted before the result.

SO;

One plan is to select 400 voters, another plan is to select 1,600 voters

If the study were conducted repeatedly (selecting different samples of people each time);

Different sample proportions would result each time, but for either sample size, they would be centered (have their mean) at the true population proportion. This is because a sample proportion deals with random experiments that cannot be predicted in advance and they are quite known to be centered about the population proportion.

5 0

2 years ago

Anna gets paid $8.75/hour working as a barista at Coffee Break. Her boss pays her $9.00/hour for creating the weekly advertiseme

vladimir1956 [14]

Answer:

i believe it would be $443.75

Step-by-step explanation:

just add 8.75 to 9.00 then multiply it by 25

8.75+9.00= 17.75 x 25= 443.75

if its a different type of equation then maybe this didn't help but i hope it does

7 0

2 years ago