98,835 divided by 790 anybody? I keep getting this wrong...

Use the counting techniques from the last chapter. A bag contains three red marbles, four green ones, one fluorescent pink one,

sergiy2304 [10]

Answer:

The probability that she gets all the red ones, given that she gets the fluorescent pink one, is P=0.0035 or 0.35%.

Step-by-step explanation:

Susan grabs four marbles at random.

We have to calculate the probabilities that he picks the 3 red ones, given that she already picked the fluorescent pink.

If it is given that the fluorescent pink is already picked, we are left with three red marbles, four green ones, two yellow ones, and four orange ones. A total of 13 marbles.

The probability that the second marble is red is 3 in 13.

The probability that the third marble is also red is 2 (the red marbles that are left) in 12 (the total amount of marbles left), as there is a picking without replacement.

The probability that the fourth marble is 1 in 11.

Then, the probability that the 3 red marbles are picked, is:

$P=P_1\cdot P_2\cdot P_3=\dfrac{3}{13}\cdot \dfrac{2}{12}\cdot \dfrac{1}{11}=\dfrac{6}{1716}=0.0035$

4 0

3 years ago

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of