Respuesta :
Answer:
[tex] P(62< X< 72)= P(X<72) -P(X<62)=0.975-0.5=0.475 [/tex]
Step-by-step explanation:
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 courtship time (minutes).
From the problem we have the mean and the standard deviation for the random variable X. [tex]E(X)=62, Sd(X)=5[/tex]
So we can assume [tex]\mu=62 , \sigma=5[/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 need values such that
[tex]P(X<\mu -\sigma)=P(X <57)=0.16[/tex]
[tex]P(X>\mu +\sigma)=P(X >67)=0.16[/tex]
[tex]P(X<\mu -2*\sigma)=P(X<52)=0.025[/tex]
[tex]P(X>\mu +2*\sigma)=P(X>72)=0.025[/tex]
[tex]P(X<\mu -3*\sigma)=P(X<47)=0.0015[/tex]
[tex]P(X>\mu +3*\sigma)=P(X>77)=0.0015[/tex]
For this case we want to find this probability:
[tex] P(62 < X< 72) [/tex]
And we can find this probability on this way:
[tex] P(62< X< 72)= P(X<72) -P(X<62) [/tex]
Since [tex] P(X>72) =0.025[/tex] by the complement rule we have that:
[tex] P(X<72) = 1-0.025 =0.975[/tex]
And [tex] P(X<62) =0.5[/tex] because for this case 62 is the mean.
So then we have this:
[tex] P(62< X< 72)= P(X<72) -P(X<62)=0.975-0.5=0.475 [/tex]
