All categories

3 years ago

I need help with a scatter plot

Mathematics

1 answer:

xeze [42]3 years ago

4 0

Answer:

Scatter plots are like graphs in which you plot points. The rate can vary. They are mainly used to compare temperatures, or weather. Also other types of data like population.

You might be interested in

I need a group of 4 functions and explain why they're functions.

Black_prince [1.1K]

Answer:

A group functions when everyone works together

5 0

3 years ago

A student wanted to construct a 95% confidence interval for the mean age of students in her statistics class. She randomly selec

VMariaS [17]

Answer:

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=19.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=9 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=9-1=8$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,8)".And we see that $t_{\alpha/2}=$

Now we have everything in order to replace into formula (1):

$19.1-3.355\frac{1.5}{\sqrt{9}}=17.42$

$19.1+3.355\frac{1.5}{\sqrt{9}}=20.78$

And the best option would be:

C. [17.42,20.78]

5 0

3 years ago

I need help asapppp!!!!!!!

monitta

Answer:=15x^2+250x

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

The Town of Hertfordshire clerk knows that 23% of dogs in the town have completed emotional support training. Hertfordshire plan

Nataly_w [17]

Answer:

95.64% probability that under 30% of the dogs are emotional support trained

Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they have completed emotional support training, or they have not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .