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2 years ago

The purpose of statistical inference is to make estimates or draw conclusions about a.

1 answer:

4 0

When we make estimates of or draw conclusions about one or more characteristics of a population based upon the sample, we are using the process of statistical inference.

To estimate this sample to sample variance or uncertainty is the goal of statistical inference.

What is the purpose of statistical inferences ?

To be able to make inferences about a population based on data from a sample is the goal of statistical inference.
The process of statistical inference involves selecting a sample, gathering data from that sample, calculating a statistic from the data, and drawing conclusions about the population from that statistic.

How is statistical inference used to draw conclusions?

Estimation and hypothesis testing are components of statistical inference (evaluating a notion about a population using a sample) (estimating the value or potential range of values of some characteristic of the population based on that of a sample).

Learn more about statistical inference

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Answer:

a) $P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

b) $P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

c) $P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816$

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=10, p=0.32)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

Part b

$P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]$

$P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211$

$P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995$

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

$P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

Part c

$P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)$

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

$P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264$

$P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218$

$P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123$

$P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816$

6 0

3 years ago

A fundraising event offered people a chance to pay $3 to go through a corn maze (shown below) in a field in the fall. The maze e

german

*100% correct answers

Question 1

A fundraising event offered people a chance to pay $3 to go through a corn maze (shown below) in a field in the fall. The maze either led to nothing or a prize of $4 (ie their $3 back and an extra $1). The participants can only move forward and when they come to a fork in the maze they are equally likely to take any one of the paths. What is the expected monetary value for each participant that goes through the maze?

-0.33

Question 2

What is the expected value of the spinner shown?

2.125

Question 3

An insurance company charges a customer $1600 per year for a particular customers auto insurance. The company has predicted that there is a 10% change the person will make a claim on the policy of $5000 (which means the insurance company would lose $3400) and 90% change that they won't make a claim. What can the insurance company on average expect to make on selling this policy?

$1100

4 0

3 years ago