please help thank you

Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118. If a recent test-taker

LuckyWell [14K]

Answer:

Probability that the student scored between 455 and 573 on the exam is 0.38292.

Step-by-step explanation:

We are given that Math scores on the SAT exam are normally distributed with a mean of 514 and a standard deviation of 118.

Let X = Math scores on the SAT exam

So, X ~ Normal( $\mu=514,\sigma^{2} =118^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean score = 514

$\sigma$ = standard deviation = 118

Now, the probability that the student scored between 455 and 573 on the exam is given by = P(455 < X < 573)

P(455 < X < 573) = P(X < 573) - P(X $\leq$ 455)

P(X < 573) = P( $\frac{X-\mu}{\sigma}$ < $\frac{573-514}{118}$ ) = P(Z < 0.50) = 0.69146

P(X $\leq$ 2.9) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{455-514}{118}$ ) = P(Z $\leq$ -0.50) = 1 - P(Z < 0.50)

= 1 - 0.69146 = 0.30854

The above probability is calculated by looking at the value of x = 0.50 in the z table which has an area of 0.69146.

Therefore, P(455 < X < 573) = 0.69146 - 0.30854 = 0.38292

Hence, probability that the student scored between 455 and 573 on the exam is 0.38292.

7 0

3 years ago

Which of the following is equivalent to 3x−4y=18?

sammy [17]

The equation which is equivalent to the equation as given in the task content is; 6x -8y = 36.

<h3>Which equation is equivalent to the equation as given in the task content?</h3>

According to the task content, it follows that the equation which is given in the task content is;

3x -4y = 18.

Hence, upon multiplication of the whole equation by 2; the resulting equation from the multiplication is;

6x -8y = 36

Ultimately, the equivalent equation is; 6x -8y = 36.

Read more on equivalent equations;

brainly.com/question/26437692

#SPJ1

8 0

1 year ago

Please help me with this. I forgot the formula for this

Lapatulllka [165]

There is the formula now just plug in the values and you will get your answer.

7 0

3 years ago

Suppose that the weights of 5400 registered female Labrador retrievers in the United States are distributed normally with a mean

Maurinko [17]

Answer:

$N= 4543$ Labrador retrievers

Step-by-step explanation:

We know that the mean $\mu$ is:

$\mu = 62.5$

and the standard deviation $\sigma$ is:

$\sigma=2.5$

The probability that a randomly selected Labrador retriever weighs less than 65 pounds is:

$P(X$

We calculate the Z-score for X =65

$Z = \frac{X-\mu}{\sigma}\\\\Z =\frac{65-62.5}{65}=1$

So

$P(X$

Looking in the table for the standard normal distribution we have to:

$P(Z$ .

Finally the amount N of Labrador retrievers that weigh less than 65 pounds is:

$N = P(X$

$N = 0.8413*5400$

$N= 4543$ Labrador retrievers

6 0

3 years ago

Help me fast ASAP pls

Aleonysh [2.5K]

Answer:

C

Step-by-step explanation:

I got it right

5 0

2 years ago

￼please help thank you

please help thank you