La mitad de un número más tres cuartos del número es 3 menos que cuatro tercios del número. Encuentre el número.

The mean weight of 10 randomly selected newborn babies at a local hospital is 7.14 lbs and the standard deviation is 0.87 lbs. A

sergejj [24]

Answer:

a) $ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035$

b) 90% of all samples of size 10 have sample means within 0.5035 of the population mean.

c) The 90% confidence interval would be given by (6.636;7.643)

d) Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

e) If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value $t_{\alpha/2}$ increase. So then the new interval would be wider than the original.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=7.14$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=0.87 represent the sample standard deviation

n=10 represent the sample size

Part a

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And the margin of error is given by:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence 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 value. The excel command would be: "=-T.INV(0.05,9)".And we see that $t_{\alpha/2}=1.83$

$ME= 1.83 \frac{0.87}{\sqrt{10}}=0.5035$

Part b

90% of all samples of size 10 have sample means within 0.5035 of the population mean.

Part c

Now we have everything in order to replace into formula (1):

$7.14-1.83\frac{0.87}{\sqrt{10}}=6.637$

$7.14+1.83\frac{0.87}{\sqrt{10}}=7.643$

So on this case the 90% confidence interval would be given by (6.636;7.643)

Part d

Yes, since the lower limit for the 90% confidence interval is higher than the value of 6.5 we can conclude that the true mean is significantly higher than 6.5 at 10% of significance. (6.636>6.5)

Part e

If we increase the confidence level that implies increase the margin of error. Since with more confidence level the value for the critical value $t_{\alpha/2}$ increase. So then the new interval would be wider than the original.

7 0

4 years ago

HELP PLEASE !!!! SHOW WORK TO ANSWER PLEASE

likoan [24]

Answer: ~14°, ~62°

Step-by-step explanation:

See image. You should be able to easily calculate the last two angles.

5 0

3 years ago

You flip a coin that is not fair, the prbability of heads on each flip is 0.7. if the coin shows heads, you draw a marble from u

Shkiper50 [21]

Solution :

P(H) = 0.7 ; P(T) = 0.3

If heads, then Urn H, 1 blue and 4 red marbles.

If tails, then Urn T , 3 blue and 1 red marbles.

a).

P ( choosing a Red marble )

= P (H) x P( Red from Urn H) + P (T) x P( Red from Urn T)

$=0.7 \times \frac{4}{5} + 0.3 \times \frac{1}{4}$

= 0.56 + 0.075

= 0.635

b). If P (B, if coin showed heads)

If heads, then marble is picked from Urn H.

Therefore,

P (Blue) $=\frac{1}{5}$

= 0.2

c). P (Tails, if marble was red)

$=P (T/R) = \frac{P(R/T)}{P(R)} \ P(T)$

Where P (R/T) = P ( red, if coin showed tails)

$=\frac{1}{4}$

= 0.25 (As Urn T is chosen)

P (R) = P (Red) = 0.635 (from part (a) )

P (T) = P (Tails) = 0.3

∴ $P(T/R) = \frac{0.25 \times 0.3}{0.635}$

= 0.118

8 0

3 years ago

Katya

alexandr402 [8]

Step-by-step explanation:

The seedling grew 1 1/6 inches or approximately 1.16 inches. 1/3 is equal to 2/6, 2/6 + 5/6 = 7/6

7/6 = 1 1/6

4 0

3 years ago

How can the Pythagorean Identity be used to find sin θ, cos θ, or tan θ and the quadrant location of the angle?

Archy [21]

First lets see the pythagorean identities

$sin^2 \Theta + cos^2 \Theta =1$

So if we have to solve for sin theta , first we move cos theta to left side and then take square root to both sides, that is

$sin^2 \Theta = 1-cos^2 \Theta => sin \theta = \pm \sqrt{1-cos^2 \Theta}$

Now we need to check the sign of sin theta

First we have to remember the sign of sin, cos , tan in the quadrants. In first quadrant , all are positive. In second quadrant, only sin and cosine are positive. In third quadrant , only tan and cot are positive and in the last quadrant , only cos and sec are positive.

So if theta is in second quadrant, then we have to positive sign but if theta is in third or fourth quadrant, then we have to use negative sign .

5 0

3 years ago