All categories

3 years ago

How is a calculation of net worth different from a day to day or month tallying of expenses

1 answer:

3 0

Net worth day 2 day not really. look in wallet. month 2 month check saving/checking account. net worth isn't done on either normally.

You might be interested in

nventing is a difficult way to make money. Only 5% of new patents earn a substantial profit. A certain city has just had30 indep

Arlecino [84]

Answer:

$P(X \geq 2) = 1-P(X$

And we can find the individual probabilities using the probability mass function and we got:

$P(X=0) = (30C0) (0.05)^0 (1-0.05)^{30-0} =0.2146$

$P(X=1) = (30C1) (0.05)^1 (1-0.05)^{30-1} = 0.3389$

And replacing we got:

$P(X \geq 2) = 1-P(X$

Step-by-step explanation:

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".

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

$X \sim Binom(n=30, p=0.05)$

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:

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

Solution to the problem

For this case we want this probability:

$P(X \geq 2)$

And we can use the complement rule and we got:

$P(X \geq 2) = 1-P(X$

And we can find the individual probabilities using the probability mass function and we got:

$P(X=0) = (30C0) (0.05)^0 (1-0.05)^{30-0} =0.2146$

$P(X=1) = (30C1) (0.05)^1 (1-0.05)^{30-1} = 0.3389$

And replacing we got:

$P(X \geq 2) = 1-P(X$

3 0

3 years ago

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon

fiasKO [112]

Answer:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.

This means that $n = 603, \pi = \frac{142}{603} = 0.2355$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694$

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

8 0

2 years ago

HELP!!!! Choose the correct answer. Which expression, when evaluated,

blagie [28]

We will check each options and verify it

(a)

$18-2m$