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Sav [38]
2 years ago
7

In a study of texting speed, 45 adults aged 18-22 were randomly chosen. Each person was asked to the following phrase exactly "H

i! What's up?" on a cell phone. The average time it took was 2.72 seconds. This was found to happen with about a 25% chance when compared to someone saying it takes less than 3 seconds to type. 1.) Identify the following: a. Population: 1. b. Sample: c. Unit/individual: d. Response variable: 2.) What type of variable is the response variable? And what is the level of measurement?
Mathematics
1 answer:
e-lub [12.9K]2 years ago
3 0

From the given data in the study of texting speed, we can identify the following

1)

a) population; is all the individuals belonging to the age group of 18-22 years

b) sample; is the selected 45 adults in the age group of 18-22 years

c) unit / individual: adults

d) Response: time taken to write "hi! what's up?"

2)

<h3>The response variable is </h3>
  • A continuous - quantitative variable.

It is a continuous variable as it can take an infinite number of possible values.

<h3>The level of measurement is</h3>
  • Measured by the ratio scale of measurement.

variables measured by the ratio scale of measurement have equal differences between scale values and equal quantitative meaning. They also have a true zero point

For more information on response variable, visit

brainly.com/question/14662435?referrer=searchResults

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b.  0.086291

c.  0.00058

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Step-by-step explanation:

a. -A normal distribution is expressed in the form X~N(mean, standard deviation).

-Let X a random variable denoting  the number of calories consumed.

-X is a is a normally distributed random variable with mean 2885 and standard deviation 651.

-This distribution is expressed as X~N(2,885, 651)

b. The probability that less than 2000 calories are consumed is calculated using the formula:

P(X

#substitute the given values in the formula to solve for P:

P(X

Hence, the probability of consuming less than 2000 calories is 0.08691

c. The proportion of customers consuming more than 5000 calories is calculated as:

P(X>x)=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(Z>\frac{5000-2885}{651})\\\\=P(z>3.2488)\\\\=1-0.99942\\\\=0.00058

Hence, the proportion of customers consuming over  5000 calories is 0.00058

d. The least amount of calories to get the award is calculated as:

1% is equivalent to a z value of 0.50399.

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0.01=P(z>\frac{\bar X-\mu}{\sigma})\\\\=P(z>\frac{\bar X-2885}{651})\\\\\1\%=0.50399 \\\\\frac{\bar X-2885}{651}=0.50399 \\\\\bar X=0.50399\times 651+2885\\\\=3213.09

Hence, the least amount of calories consumed to qualify for the award is 3213.10 calories.

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