Find the distance between the points. (0, 4) and (3, 0)

One of the authors with too much time on his hands weighed each chocolate candy in a bag of 463 plain chocolate candies. Complet

wel

Answer:

a. One of the chocolate candies weighed 0.776 gram and it was heavier than 25 of the other chocolate candies. What is the percentile of this particular value?

It is the 5th percentile.

b. One of the chocolate candies weighed 0.876 gram and it was heavier than 329 of the other chocolate candies. What is the percentile of this particular value?

It is the 71th percentile.

c. One of the chocolate candies weighed 0.856 gram and it was heavier than 229 of the other chocolate candies. What is the percentile of this particular value?

It is the 49th percentile.

Step-by-step explanation:

So when we are talking about percentile, what it means is the point below a certain score or percentage lies, or the rank.

The formula is simple:

Let's call percentile pth, pth= $\frac{p}{100}*(n+1)$

where p is the position in the data set you want to evaluate (in statistical analysis, these are called quartiles, and they are useful for evaluating groups, so 3 quartiles would divide sample into 25th, 50th and 75th, four quartiles would be 20th, 40th, 60th, 80th, and so on).

and n is the number of elements in the set

For this problem there are no quartiles specified, but we can then work with the definition of percentile, meaning, if you are the 5th percentile, then 95% of the group are above you.

a. So 0.776 was heavier than 25 others, meaning it is in the 26th position out of 463. But we need that in relation to 100, so $\frac{26}{463}$ is 0.0562 or about 5/100, (rounded down). That would mean that the percentile is:

pth= $\frac{26}{463}$ *100=5.62 or 5, rounded down

b. 0.876 was heavier than 329 others, meaning it is in the 330th position out of 463. In relation to 100, so $\frac{330}{463}$ is 0.7127 or about 71/100, (rounded down). That would mean that the percentile is:

pth= $\frac{330}{463}$ *100=71.27 or 71, rounded down

c. 0.856 was heavier than 229 others, meaning it is in the 230th position out of 463. In relation to 100, so $\frac{230}{463}$ is 0.4967 or about 49/100, (rounded down). That would mean that the percentile is:

pth= $\frac{230}{463}$ *100=49.67 or 49, rounded down

3 0

4 years ago

The heights of a certain type of tree are approximately normally distributed with a mean height

Makovka662 [10]

Let the height of the trees are h1,h2 and h3,
Then,we know that
Mean height=(h1+h2+h3)/3

4 0

4 years ago

The germination rate is the rate at which plants begin to grow after the seed is planted.

chubhunter [2.5K]

Answer:

$z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59$

$p_v =P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%

Step-by-step explanation:

Data given and notation

n=15 represent the random sample taken

X=7 represent the number of seeds germinated

$\hat p=\frac{7}{15}=0.467$ estimated proportion of seeds germinated

$p_o=0.9$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion of germinated seeds is less than 0.9 or 90%.:

Null hypothesis: $p\geq 0.9$

Alternative hypothesis: $p < 0.9$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%

4 0

3 years ago

Which equations are in standard form? Check all that apply

NemiM [27]

Standard form is y=mx+b
y = 2x + 5
y = x – 9

8 0

3 years ago

Evaluate each expression. 18^-4 x 6^7

rosijanka [135]

Answer:

8/3

Step-by-step explanation: