Identifying Rays and Segments

A website reports that 70% of its users are from outside a certain country, and 60% of its users logon the website every day. Su

JulijaS [17]

Answer:

The value is $P (I | L ) = 0.63$

The probability has increased

Step-by-step explanation:

From the question we are told that

The percentage that are from outside the country is $P(O) = 0.70$

The percentage that logs on everyday is $P(L) = 0.60$

The percentage that logs on everyday that are from the inside the country is $P(L | I) = 0.80$

Generally using Bayes' Rule the probability that a person is from the country given that he logs on the website every day is mathematically represented as

$P (I | L ) = \frac{P(I)* P(L|I)}{ P(O) *P(L|O) + P(I) *P(L|I) }$

Where $P(I)$ is the percentage that are from inside that country which is mathematically represented as

$P(I) = 1 - P(O)$

$P(I) = 1 - 0.70$

$P(I) = 0.30$

And $P(L| O)$ is percentage that logs on everyday that are from the outside the country which is evaluated as

$P(L| O) = 1- P(L| I)$

$P(L| O) = 1- 0.80$

$P(L| O) = 0.20$

$P (I | L ) = \frac{ 0.3* 0.80 }{ 0.7 *0.20 + 0.3 * 0.8 }$

$P (I | L ) = 0.63$

Given that the percentage that are from inside that country is $P(I) = 0.30$

and that the probability that a person is from the country given that he logs on the website every day is $P (I | L ) = 0.63$

We see that the additional information increased the probability

5 0

2 years ago

Use an inequality symbol (<, >, =, ≥) to compare −17 ____ −1

o-na [289]

A is the right answer

God Bless

5 0

2 years ago

Eric throws a biased coin 10 times. He gets 3 tails. Sue throw the same coin 50 times. She gets 20 tails. Aadi is going to throw

NISA [10]

Answer:

(1) Correct option (A).

(2) The probability that Aadi will get Tails is 0.40.

Step-by-step explanation:

The information provided is:

Eric throws a biased coin 10 times. He gets 3 tails.
Sue throw the same coin 50 times. She gets 20 tails.

The probability of tail in both cases is:

$P(T|E)=\frac{3}{10}=0.30$

$P(T|S)=\frac{20}{50}=0.40$

Here,

P (T|E) implies the probability of tail in case of Eric's experiment.

P (T|S) implies the probability of tail in case of Sue's experiment.

(1)

Now, it is given that Aadi is going to throw the coin once.

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

In this case we need to compute the probability of Aadi getting Tails in a single toss.

As Sue uses a larger number of trials in the experiment, i.e. n = 50 > 30 times, according to the Central limit theorem, Sue's estimate is best because she throws it .

Thus, the correct option is (A).

(2)

As explained in the first part that Sue's estimate is best for getting a tail, the probability that Aadi will get Tails when he tosses the coin once is:

$P(\text{Aadi will get Tails})=P(T|A)$

$=P(T|S)\\\\=0.40$

Thus, the probability that Aadi will get Tails is 0.40.

8 0

2 years ago

Read 2 more answers

A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is th

GenaCL600 [577]

Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. 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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .