Respuesta :
Answer:
a) For this case we can use this:
[tex]P(X<\mu -3*\sigma)=P(X<32.1)=0.0015[/tex]
[tex]P(X>\mu +3*\sigma)=P(X>46.5)=0.0015[/tex]
So the range of lengths would be 32.1 and 46.5 since on the middle of these two values we have 1-0.0015-0.0015=0.997 or 99.7 % of the values.
b) [tex] P(X>36.9) [/tex]
And using the complement rule we have:
[tex]P(X>36.9)= 1-P(X<36.9) = 1-0.16= 0.84[/tex]
Step-by-step explanation:
Previous concepts
The empirical rule, also known as three-sigma rule or 68-95-99.7 rule, "is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ)".
Let X the random variable who represent the variable of interest.
From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=39.3, Sd(X)=2.4[/tex]
So we can assume [tex]\mu=39.3 , \sigma=2.4[/tex]
On this case in order to check if the random variable X follows a normal distribution we can use the empirical rule that states the following:
• The probability of obtain values within one deviation from the mean is 0.68
• The probability of obtain values within two deviation's from the mean is 0.95
• The probability of obtain values within three deviation's from the mean is 0.997
So we have the following probabilities using this rule:
[tex]P(X<\mu -\sigma)=P(X <36.9)=0.16[/tex]
[tex]P(X>\mu +\sigma)=P(X >41.7)=0.16[/tex]
[tex]P(X<\mu -2*\sigma)P(X<34.5)=0.025[/tex]
[tex]P(X>\mu +2*\sigma)P(X>44.1)=0.025[/tex]
[tex]P(X<\mu -3*\sigma)=P(X<32.1)=0.0015[/tex]
[tex]P(X>\mu +3*\sigma)=P(X>46.5)=0.0015[/tex]
Part a
For this case we can use this:
[tex]P(X<\mu -3*\sigma)=P(X<32.1)=0.0015[/tex]
[tex]P(X>\mu +3*\sigma)=P(X>46.5)=0.0015[/tex]
So the range of lengths would be 32.1 and 46.5 since on the middle of these two values we have 1-0.0015-0.0015=0.997 or 99.7 % of the values.
Part b
For this casee we want this probability:
[tex] P(X>36.9) [/tex]
And using the complement rule we have:
[tex]P(X>36.9)= 1-P(X<36.9) = 1-0.16= 0.84[/tex]
The range of the lengths would be 32.1 and 46.5. Since on the middle of two values we have [tex]1-0.0015 - 0.0015 = 0.997[/tex] or 99.7% of the values. The complement rule we have [tex]\rm P(X >36.9) = 1 - P(X<36.9) = 1 - 0.16 = 0.84[/tex]
What is normal a distribution?
It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.
Given
Assume the upper arm length of males over 20 years old in the United States is approximately Normal with a mean of 39.3 centimeters (cm) and a standard deviation of 2.4 cm.
Let X be the random variable that represents the variable of interest.
Mean = 39.3
Standard deviation = 2.4
If the random variable X follows a normal distribution we can use the empirical rule that states the following;
- The probability of obtaining values within one deviation from the mean is 0.68.
- The probability of obtaining values within two deviations from the mean is 0.95.
- The probability of obtaining values within three deviations from the mean is 0.997.
So we have the following probabilities using the rule:
[tex]\rm P(X< \mu - \sigma) = P(X< 36.9) = 0.16\\\\P(X> \mu - \sigma) = P(X> 41.7) = 0.16\\\\P(X< \mu -2* \sigma) = P(X< 34.5) = 0.25\\\\P(X> \mu -2* \sigma) = P(X> 44.1) = 0.25\\\\P(X< \mu - 3* \sigma) = P(X< 32.1) = 0.0015\\\\P(X> \mu - 3* \sigma) = P(X> 46.5) = 0.0015\\\\[/tex]
Part A. For this case, we can use this.
[tex]\rm P(X< \mu - 3* \sigma) = P(X< 32.1) = 0.0015\\\\P(X> \mu - 3* \sigma) = P(X> 46.5) = 0.0015\\\\[/tex]
So, the range of the lengths would be 32.1 and 46.5. Since on the middle of two values we have
[tex]1-0.0015 - 0.0015 = 0.997[/tex] or 99.7% of the values.
Part B. For this case we want this probability:
[tex]\rm P(X>36.9)[/tex]
And using the complement rule we have
[tex]\rm P(X >36.9) = 1 - P(X<36.9) = 1 - 0.16 = 0.84[/tex]
More about the normal distribution link is given below.
https://brainly.com/question/12421652
