In box kernel density estimation, . a. None of the options b. The histogram is decentralized over several data points. c. The histogram is decentralized. d. The histogram is center

The correct word for the blank is the histogram is centralized at many points.

The non-parametric way of assessing or estimating the density function of the random variable is known as the Kernel density estimation.

Kernel density estimation is referred to the fundamental data smoothing problem that involves population inferences that are obtained from a finite data sample.

It is also known as the Parzen–Rosenblatt window approach in some domains for example signal processing and econometrics, after Emanuel Parzen and Murray Rosenblatt, who are usually credited with independently developing it in its current form.

When making use of a naive Bayes classifier, one of the most well-known applications of kernel density estimation is calculating the class-conditional marginal densities of data, which can enhance prediction accuracy.  

Therefore, the correct option is d.

To know more about the  box kernel density estimation, refer to the link below:

link

D. The histogram is centered over several data points.

Step-by-step explanation:

The Box density Estimation is used when given several values of data plotted as data points, to generate a smooth curve. The histogram in the box kernel estimate is centered over several data points. Blocks are now placed on each of the data points, and this helps to eliminate the histogram's reliance on the endpoints. It results in some sort of convergence.

The major differences between the kernel density estimation and histogram is that there are no endpoints, it is smooth, and relies largely on bandwidth.

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step-by-step explanation:

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.