1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
shusha [124]
3 years ago
6

Can you solve them with points to plot thank you

Mathematics
1 answer:
BigorU [14]3 years ago
8 0

I have solved the first one. Hope this helps

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
Elena has a washing machine with a 20 gallon capacity. She always uses a 20 gram measure of soap for every 10 gallons of water.
baherus [9]

Answer:

For \[20\] minutes Elena add water to correct the error.

Step-by-step explanation:

The capacity of the washing machine is \[20\] gallons.

She uses \[20\] gram soap for \[10\] gallon of water

Her son by mistake dissolved \[60\] grams of soap to \[10\] gallons of water.

\[60\]-gram soap is right for \[30\] gallons of water, thus \[20\] gallon of water must be added more.

She pore water \[2\] gallons per minute thus, \[20\] gallons of water will take\[10\]minutes.

Learn more about equations : brainly.com/question/20727602?referrer=searchResults

6 0
2 years ago
Interior and Exterior Triangle Angles
k0ka [10]
Answer: 20.2
Add all the angles and set it equal to 180
Combine like terms
20x-2=180
X= 9.1

The equation for v is 2x+2
Replace x with 9.1 and then solve
3 0
3 years ago
What is the perimeter and area of length is 13 inches in width is 9.5 inches
love history [14]

Answer:

perimeter = 45 in.

area = 123.5 in.^2

Step-by-step explanation:

I assume this is a rectangle.

L = 13 in.

W = 9.5 in.

perimeter = 2(L + W) = 2(13 in. + 9.5 in.) = 2(22.5 in.) = 45 in.

area = LW = 13 in. * 9.5 in. = 123.5 in.^2

7 0
3 years ago
Read 2 more answers
Find, leave your answers in terms of pie,
Yuri [45]
Area of square = 10x10
= 100
area of circle = pi(r)^2
= pi (5)^2
=25pi

area of shaded = 100 -25pi

circumference of circle = 2pi (r)
= 2pi(5)
= 10pi

perimeters of shaded = 10pi + 10 + 10
= 10pi +20
8 0
2 years ago
Other questions:
  • One more question...
    9·1 answer
  • What percent would be equivalent to \frac{8}{10} 8 10 ?
    9·1 answer
  • Aimee and Ben are purchasing a condominium and are financing $610,000. The mortgage is a 20-year 3/1 ARM at 4.15% with a cap str
    13·1 answer
  • Angle Pairs Introduction
    15·1 answer
  • What is the exact area of a circle having diameter 7 in.?
    11·1 answer
  • Y−3=2(x+1)<br> Complete the missing value in the solution to the equation.<br> (−1,
    10·2 answers
  • Which equation has solution x = -3?
    7·1 answer
  • Help me with this plsssss
    6·2 answers
  • Please please help!!
    7·1 answer
  • Use the muliplier method to increase 88 pound by 14% show your working
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!