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Talja [164]
2 years ago
10

Ocean waves move in parallel lines toward the shore. The figure shows the path that a windsurfer takes

Mathematics
1 answer:
Minchanka [31]2 years ago
4 0

Answer:

nopw

Step-by-step explanation:

You might be interested in
Problem: The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72
Lisa [10]

Answer:

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

1) 0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2) 0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3) 0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4) 0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

Step-by-step explanation:

To solve these questions, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The height, X, of all 3-year-old females is approximately normally distributed with mean 38.72 inches and standard deviation 3.17 inches.

This means that \mu = 38.72, \sigma = 3.17

Sample of 10:

This means that n = 10, s = \frac{3.17}{\sqrt{10}}

Compute the probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

This is 1 subtracted by the p-value of Z when X = 40. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{40 - 38.72}{\frac{3.17}{\sqrt{10}}}

Z = 1.28

Z = 1.28 has a p-value of 0.8997

1 - 0.8997 = 0.1003

0.1003 = 10.03% probability that a simple random sample of size n= 10 results in a sample mean greater than 40 inches.

Gestation periods:

\mu = 266, \sigma = 16

1. What is the probability a randomly selected pregnancy lasts less than 260 days?

This is the p-value of Z when X = 260. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{260 -  266}{16}

Z = -0.375

Z = -0.375 has a p-value of 0.3539.

0.3539 = 35.39% probability a randomly selected pregnancy lasts less than 260 days.

2. What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less?

Now n = 20, so:

Z = \frac{X - \mu}{s}

Z = \frac{260 - 266}{\frac{16}{\sqrt{20}}}

Z = -1.68

Z = -1.68 has a p-value of 0.0465.

0.0465 = 4.65% probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less.

3. What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less?

Now n = 50, so:

Z = \frac{X - \mu}{s}

Z = \frac{260 - 266}{\frac{16}{\sqrt{50}}}

Z = -2.65

Z = -2.65 has a p-value of 0.0040.

0.004 = 0.4% probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less.

4. What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

Sample of size 15 means that n = 15. This probability is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256.

X = 276

Z = \frac{X - \mu}{s}

Z = \frac{276 - 266}{\frac{16}{\sqrt{15}}}

Z = 2.42

Z = 2.42 has a p-value of 0.9922.

X = 256

Z = \frac{X - \mu}{s}

Z = \frac{256 - 266}{\frac{16}{\sqrt{15}}}

Z = -2.42

Z = -2.42 has a p-value of 0.0078.

0.9922 - 0.0078 = 0.9844

0.9844 = 98.44% probability a random sample of size 15 will have a mean gestation period within 10 days of the mean.

8 0
3 years ago
A bag contains only red, green, brown and yellow marbles.
Mila [183]

Answer:

nikky ggeibf theke jmbhknfyk be s you know vfub be fj set u in vy d so o ka vgu

8 0
2 years ago
1.3 billion people live on?
Lunna [17]

Answer:

1.25

Step-by-step explanation:

In India

4 0
3 years ago
Read 2 more answers
How many places are in fifty million without writing the number in standard form?
Natali [406]

In order to answer this problem, you need to remember the place – value chart since you don't want to convert the number to its standard form. The number "fifty million" has nine places. Enumerating from the left would be: 

<span>- Ones - Tens - Hundreds - Thousands - Tens thousands - Hundreds thousands - Millions - T<span>ens millions</span></span>
8 0
3 years ago
A movie theater charges $15 for standard viewing, $20 for 3D viewing, and $35 for Dinner and a
Ludmilka [50]

The quantity of each type of seats sold are as follows:

  • Standard viewing = 2000

  • 3D seats = 800

  • Movie and a dinner seat = 200

According to the question,there are four times as many 3D, x seats as Dinner and a Movie, y seats.

That is, x = 4y

Also, total seats

= (x) + (y) + (z) = 3000...….............eqn(1)

Also, If the theater brings in $53,000 when tickets to all 3000 seats are sold.

  • 20x + 35y + 15z = 53000...........eqn(2)

By substituting 4y for x in equations 1 and 2; we have;

<em>5y + z = 3000</em>..…........eqn(3) and

<em>115y + 15z = 53000</em>.........eqn(4)

By solving equations 3 and 4 simultaneously; we have;

y = 200 and z = 2000

and since x = 4y

x = 800

The quantity of each type of seats sold is as follows:

  • Standard viewing = 2000

  • 3D seats = 800

  • Movie and a dinner seat = 200

Read more:

brainly.com/question/12413726

5 0
3 years ago
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