Convert the following units and rates

Assume that SAT scores are normally distributed with mean 1518 and standard deviation 325. Round your answers to 4 decimal place

Katyanochek1 [597]

Answer:

a. 0.2898

b. 0.0218

c. 0.1210

d. 0.1515

e. This is because the population is normally distributed.

Step-by-step explanation:

Assume that SAT scores are normally distributed with mean 1518 and standard deviation 325. Round your answers to 4 decimal places

We are using the z score formula when random samples

This is given as:

z = (x-μ)/σ/√n

where x is the raw score

μ is the population mean

σ is the population standard deviation.

n is the random number of samples

a.If 100 SAT scores are randomly selected, find the probability that they have a mean less than 1500.

For x = 1500, n = 100

z = 1500 - 1518/325/√100

z = -18/325/10

z = -18/32.5

z = -0.55385

Probability value from Z-Table:

P(x<1500) = 0.28984

Approximately = 0.2898

b. If 64 SAT scores are randomly selected, find the probability that they have a mean greater than 1600

For x = 1600, n = 64

= z = 1600 - 1518/325/√64.

z= 1600 - 1518 /325/8

z = 2.01846

Probability value from Z-Table:

P(x<1600) = 0.97823

P(x>1600) = 1 - P(x<1600) = 0.021772

Approximately = 0.0218

c. If 25 SAT scores are randomly selected, find the probability that they have a mean between 1550 and 1575

For x = 1550, n = 25

z = 1550 - 1518/325/√25

z = 1550 - 1518/325/5

z = 1550 - 1518/65

= 0.49231

Probability value from Z-Table:

P(x = 1550) = 0.68875

For x = 1575 , n = 25

z = 1575 - 1518/325/√25

z = 1575 - 1518/325/5

z = 1575 - 1518/65

z = 0.87692

Probability value from Z-Table:

P(x=1575) = 0.80974

The probability that they have a mean between 1550 and 1575

P(x = 1575) - P(x = 1550)

= 0.80974 - 0.68875

= 0.12099

Approximately = 0.1210

d. If 16 SAT scores are randomly selected, find the probability that they have a mean between 1440 and 1480

For x = 1440, n = 16

z = 1440 - 1518/325/√16

= -0.96

Probability value from Z-Table:

P(x = 1440) = 0.16853

For x = 1480, n = 16

z = 1480 - 1518/325/√16

=-0.46769

Probability value from Z-Table:

P(x = 1480) = 0.32

The probability that they have a mean between 1440 and 1480

P(x = 1480) - P(x = 1440)

= 0.32 - 0.16853

= 0.15147

Approximately = 0.1515

e. In part c and part d, why can the central limit theorem be used even though the sample size does not exceed 30?

The central theorem can be used even though the sample size does not exceed 30 because the population is normally distributed.

6 0

3 years ago

Can somebody please help me with this question?

Aliun [14]

Answer:

120, 114, 126

Step-by-step explanation:

Since these are all exterior angles, they will all always add up to 360°. Since we know this, you can set up an equation and it would be x+(x+6)+114=360. When added you get 2x+120=360. Now subtract 120 on each side to get 2x equals 240. Then you divide by 2 and get that x is 120. Now all you have to do is plug x into the equations for the angles so they would be 120, 114, and 120+6 so 126. Now you have 120,114,and 126 as your angles. To check, just add all the angle measurements up and it should equal 360.

4 0

3 years ago

Read 2 more answers

15. Madison invested in a certificate of deposit for 4 years at a 6% interest rate. At the end of the 4 years, the value

Reptile [31]

Answer:

2455.49

Step-by-step explanation:

$3100=x(1+.06)^4\\x=2455.490356$

6 0

2 years ago

An automobile company hired an advertising agency to survey its customers to find out which brand of tire they preferred for the

alexandr1967 [171]

A:the mean and that’s all I know ahahahahahahahhbbahhhh

7 0

3 years ago

Read 2 more answers

Find derivative problem<br> Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0

3 years ago