All categories

3 years ago

What is the probability of tossing two Fair coins and having exactly one land Tails side up

2 answers:

8 0

It would be 1/2 because there are 4 different possibilities of the two coins landing and exactly 2 of them have exactly only one side up thus concluding in the answer being 50%.

OlgaM077 [116]3 years ago

5 0

1/4 the probability of the 1st coin is 1/2 for landing on tails and for the second coin, the probability is 1/2 for it not to land on tails so u do 1/2* 1/2 and u get 1/4

You might be interested in

A race car driving under the caution flag at 40 feet per second begins to accelerate at a constant rate after

vitfil [10]

Chelsea vs. Atlético Madrid, Champions League: Confirmed lineups; how to watch

7 0

3 years ago

2/7 find the reciprocal

Mars2501 [29]

7/2 is the reciprocal of 2/7

7 0

3 years ago

Read 2 more answers

The age of some lecturers are 42,54,50,54,50,42,46,46,48 and 48 calculate the mean age and standard deviation

rjkz [21]

Answer:

Mean age: 48

Standard deviation: 4

Step-by-step explanation:

a) Mean

The formula for Mean = Sum of terms/ Number of terms

Number of terms

= 42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48+ 48/ 10

= 480/10

= 48

The mean age is 48

b) Standard deviation

The formula for Standard deviation =

√(x - Mean)²/n

Where n = number of terms

Standard deviation =

√[(42 - 48)² + (54 - 48)² + (50 - 48)² +(54 - 48)² + (50 - 48)² +(42 - 48)² + (46 - 48)² + (46 - 48)² + (48 - 48)² + (48 - 48)² / 10]

= √-6² + 6² + 2² + 6² + 2² + -6² + -2² + -2² + 0² + 0²/10

=√36 + 36 + 4 + 36 + 4 + 36 + 4 + 4 + 0 + 0/ 10

=√160/10

= √16

= 4

The standard deviation of the ages is 4

3 0

4 years ago

Mathhrjirjrrggghuughihihughhhg

marusya05 [52]

Answer:

jhliyuglkiygedrloqayfgwelJUQHAUGWDkutygsk,Hgkuyt

Step-by-step explanation:

3 0

3 years ago

83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi

arlik [135]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

Step-by-step explanation:

Information given

$\bar X=32.1$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom given by:

$df=n-1=8-1=7$

The confidence level is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

$\chi^2_{\alpha/2}=104.139$

$\chi^2_{1- \alpha/2}=62.132$

The confidence interval is given by:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.525 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.127 \leq \sigma \leq 2.757$

5 0

3 years ago