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ollegr [7]
2 years ago
12

Csikszentmihalyi & Figurski (1982) conducted a study to determine the degree to which people find out about themselves throu

gh the process of introspection (thinking about themselves). They gave people beepers and asked them to write down what they were thinking about when the beepers went off randomly throughout the day. The results indicated that:
Mathematics
1 answer:
NARA [144]2 years ago
5 0

Answer:

We think about ourselves way less than we think 8% of the time.

Step-by-step explanation:

In a self-awareness: introspection research study conducted by Mihaly Csikszentmihalyi and Thomas Figurski in 1982, to determine the degree to which people find out about themselves through the process of introspection (thinking about themselves).

The research involved an arrangement where participants were given beepers and asked to write down what they were thinking about when the beepers went off randomly throughout the day. The results indicated that "We think about ourselves way less than we think 8% of the time."

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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
2 years ago
Simplify the polynomial 3x^2+5x-5x^2-4x+5-2
zepelin [54]
Hope this helps tried to color code it

7 0
3 years ago
Write that number as a decimal?
pshichka [43]
Hmm a number as. a decimal is 1/2=0.50
5 0
3 years ago
A survey was given asking whether people like dogs and/or cats. 153 said they like dogs 191 said they like cats 56 said they lik
lapo4ka [179]

Answer:

There were 362 people surveyed

Step-by-step explanation:

* <em>Lets explain how to solve the problem</em>

- A survey was given asking whether people like dogs and/or cats

# 153 said they like dogs

# 191 said they like cats

# 56 said they liked both cats and dogs

# 74 said they don't like cats or dogs

- We need to find how many people were surveyed

* <em>Lets find the number of who likes dogs only and the number who</em>

<em>  likes cats only </em>

- The number who likes dog only (cats only) is the difference between

 the number who likes dogs (cats) and the number who likes both

∴ The number who likes dogs only = 153 - 56 = 97

∴ The number who likes cats only = 191 - 56 = 135

- The total number of people were surveyed is the sum of the number

  who likes cats only and who likes dogs only and who likes both and

  who doesn't like both

∵ The number who doesn't like dogs or cats is 74

∴ The total number = 97 + 135 + 56 + 74 = 362

* There were 362 people surveyed

7 0
3 years ago
Help me with my math please!!!!
astra-53 [7]

Look up "Everything You Need To Know About Math In One Big Fat Notebook pdf." It's the best thing I've ever been given, I have it with me in math class all the time and I've aced every test. I have it with me right now and it has everything I've ever been taught about math in it so it might help you.

7 0
3 years ago
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