The lengths of adult males' hands are normally distributed with mean 191 mm and standard deviation is 7 mm. Suppose that 49 individuals are randomly chosen. Round all answers to 4 where possible. a. What is the distribution of ? N b. For the group of 49, find the probability that the average hand length is less than 192. c. Find the first quartile for the average adult male hand length for this sample size. Round to 2 decimal places. d. For part b), is the assumption that the distribution is normal necessary? Yes No
The lengths of adult males' hands are normally distributed with mean 191 mm and standard deviation is 7 mm. Suppose that 49 individuals are randomly chosen. Round all answers to 4 where possible. a. What is the distribution of ? N b. For the group of 49, find the probability that the average hand length is less than 192. c. Find the first quartile for the average adult male hand length for this sample size. Round to 2 decimal places. d. For part b), is the assumption that the distribution is normal necessary? Yes No
MATLAB: An Introduction with Applications
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![**Understanding Normal Distribution of Adult Male Hand Lengths**
The lengths of adult males' hands follow a normal distribution with a mean of 191 mm and a standard deviation of 7 mm. We consider a scenario where 49 individuals are randomly chosen. The calculations should be accurate to four decimal places where possible.
---
**a. Distribution Characteristics**
_What is the distribution of sample means?_
Given the normal distribution of the population, the distribution of the sample mean hand length (\(\overline{X}\)) for a sample size of 49 can be described by:
\[ \overline{X} \sim N \left( \mu = 191, \frac{\sigma}{\sqrt{n}} \right) \]
Here, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
Answer:
\[ \overline{X} \sim N \left( 191, \frac{7}{\sqrt{49}} \right) \]
\[ \overline{X} \sim N \left( 191, 1 \right) \]
---
**b. Probability Calculation**
_For the group of 49, find the probability that the average hand length is less than 192 mm._
To find this probability, we use the cumulative distribution function (CDF) of the normal distribution:
\[ P(\overline{X} < 192) = P\left(Z < \frac{192 - 191}{1}\right) \]
\[ P(\overline{X} < 192) = P(Z < 1) \]
Using the standard normal table or a calculator,
\[ P(Z < 1) \approx 0.8413 \]
Thus, the probability is approximately 0.8413.
---
**c. First Quartile Determination**
_Find the first quartile for the average adult male hand length for this sample size. Round to 2 decimal places._
The first quartile (Q1) in a normal distribution is the 25th percentile, which corresponds to a Z-score of approximately -0.6745.
\[ Q1 = \mu + Z \cdot \sigma \]
\[ Q1 = 191 + (-0.6745) \cdot 1 \]
\[ Q1 = 191 - 0.6745 \]
\[ Q1 \approx 190.33 \]
---
**d.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc260e632-e4e2-45c7-853f-2d07e95220d0%2F8e699f09-76a0-458f-a1b6-fa4d6618872f%2F25p7av_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Normal Distribution of Adult Male Hand Lengths**
The lengths of adult males' hands follow a normal distribution with a mean of 191 mm and a standard deviation of 7 mm. We consider a scenario where 49 individuals are randomly chosen. The calculations should be accurate to four decimal places where possible.
---
**a. Distribution Characteristics**
_What is the distribution of sample means?_
Given the normal distribution of the population, the distribution of the sample mean hand length (\(\overline{X}\)) for a sample size of 49 can be described by:
\[ \overline{X} \sim N \left( \mu = 191, \frac{\sigma}{\sqrt{n}} \right) \]
Here, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
Answer:
\[ \overline{X} \sim N \left( 191, \frac{7}{\sqrt{49}} \right) \]
\[ \overline{X} \sim N \left( 191, 1 \right) \]
---
**b. Probability Calculation**
_For the group of 49, find the probability that the average hand length is less than 192 mm._
To find this probability, we use the cumulative distribution function (CDF) of the normal distribution:
\[ P(\overline{X} < 192) = P\left(Z < \frac{192 - 191}{1}\right) \]
\[ P(\overline{X} < 192) = P(Z < 1) \]
Using the standard normal table or a calculator,
\[ P(Z < 1) \approx 0.8413 \]
Thus, the probability is approximately 0.8413.
---
**c. First Quartile Determination**
_Find the first quartile for the average adult male hand length for this sample size. Round to 2 decimal places._
The first quartile (Q1) in a normal distribution is the 25th percentile, which corresponds to a Z-score of approximately -0.6745.
\[ Q1 = \mu + Z \cdot \sigma \]
\[ Q1 = 191 + (-0.6745) \cdot 1 \]
\[ Q1 = 191 - 0.6745 \]
\[ Q1 \approx 190.33 \]
---
**d.
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