Example 6.11In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).a. Calculate the interquartile range (IQR).b. Forty percent of the smartphone users from 13 to 55+ are at least what age?Solutiona.IQR = Q3 - Q1Calculate Q3 = 75th percentile and Q1 = 25th percentile.invNorm(0.75,36.9,13.9) = Q3 = 46.2754invNorm(0.25,36.9,13.9) = Q1 = 27.5246IQR = Q3 - Q1 = 18.8bFind k where P(x ≥ k) = 0.40 ("At least" translates to "greater than or equal to.")0.40 = the area to the right.Area to the left = 1 - 0.40 = 0.60.The area to the left of k = 0.60.invNorm(0.60,36.9,13.9) = 40.4215.k = 40.4.Forty percent of the smartphone users from 13 to 55+ are at least 40.4 years. 6.9 The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.- Lower value = ______- Upper value = ______Use the TI-84 Distribution function to calculate \( P(66 < x < 70) \).[Image: Normal distribution curve]After reading through Example 6.10 (pages 376 – 377)...**Try It (page 377)**6.10 Use the information from Example 6.10 to answer the following questions.a. Find the 30th percentile, and interpret it in a complete sentence. - **Step 1** Shade the approximate area 0.3 in the diagram - **Step 2** Use the `3:invNorm(` function: - Area to the left: ______ - \(\mu\): _______ - \(\sigma\): _______ **Note for those with TI-84 Plus CE:** can choose WHERE in the distribution the desired area is – left tail, center or right tail.b. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27?[Image: Normal distribution curve]After reading through Example 6.11 (page 377)...**Try It (page 377)**6.11 Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean \(\mu = 81\) points and standard deviation \(\sigma = 15\) points.a. Calculate the first- and third-quartile scores for this exam.b. The middle 50% of the exam scores are between what two values?

6.11) Twi thousand students took an exam. The scores on the exam have an approximate normal…

Answered: After reading through Example 6.11…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Example 6.11

In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).

a. Calculate the interquartile range (IQR).

b. Forty percent of the smartphone users from 13 to 55+ are at least what age?

Solution

IQR = Q₃ - Q₁
Calculate Q₃ = 75^th percentile and Q₁ = 25^th percentile.
invNorm(0.75,36.9,13.9) = Q₃ = 46.2754
invNorm(0.25,36.9,13.9) = Q₁ = 27.5246
IQR = Q₃ - Q₁ = 18.8

Find k where P(x ≥ k) = 0.40 ("At least" translates to "greater than or equal to.")
0.40 = the area to the right.
Area to the left = 1 - 0.40 = 0.60.
The area to the left of k = 0.60.
invNorm(0.60,36.9,13.9) = 40.4215.
k = 40.4.
Forty percent of the smartphone users from 13 to 55+ are at least 40.4 years.

6.9 The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.

- Lower value = ______
- Upper value = ______

Use the TI-84 Distribution function to calculate \( P(66 < x < 70) \).

[Image: Normal distribution curve]

After reading through Example 6.10 (pages 376 – 377)...

**Try It (page 377)**

6.10 Use the information from Example 6.10 to answer the following questions.

a. Find the 30th percentile, and interpret it in a complete sentence.

- **Step 1** Shade the approximate area 0.3 in the diagram
- **Step 2** Use the `3:invNorm(` function:
- Area to the left: ______
- \(\mu\): _______
- \(\sigma\): _______

**Note for those with TI-84 Plus CE:** can choose WHERE in the distribution the desired area is – left tail, center or right tail.

b. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27?

[Image: Normal distribution curve]

After reading through Example 6.11 (page 377)...

**Try It (page 377)**

6.11 Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean \(\mu = 81\) points and standard deviation \(\sigma = 15\) points.

a. Calculate the first- and third-quartile scores for this exam.

b. The middle 50% of the exam scores are between what two values?

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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