Respuesta :
Given that in a national highway Traffic Safety Administration (NHTSA) report, data provided to the NHTSA by Goodyear stated that the mean tread life of a properly inflated automobile tires is 45,000 miles. Suppose that the current distribution of tread life of properly inflated automobile tires is normally distributed with mean of 45,000 miles and a standard deviation of 2360 miles.
Part A:
Find the probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is between two numbers, a and b is given by:
[tex]P(a \ \textless \ X \ \textless \ b) = P(X \ \textless \ b) - P(X \ \textless \ a) \\ \\ P\left(z\ \textless \ \frac{b-\mu}{\sigma} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles is given by:
[tex]P(42,000 \ \textless \ X \ \textless \ 46,000) = P(X \ \textless \ 46,000) - P(X \ \textless \ 42,000) \\ \\ P\left(z\ \textless \ \frac{46,000-45,000}{2,360} \right)-P\left(z\ \textless \ \frac{42,000-45,000}{2,360} \right) \\ \\ =P(0.4237)-P(-1.271)=0.66412-0.10183=\bold{0.5623}[/tex]
b. Find the probability that randomly selected automobile tire has a tread life of more than 50,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, a, is given by:
[tex]P(X \ \textgreater \ a) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of more than 50,000 miles is given by:
[tex]P(X \ \textgreater \ 50,000) = 1 - P(X \ \textless \ 50,000) \\ \\ =1-P\left(z\ \textless \ \frac{50,000-45,000}{2,360} \right)=1-P(z\ \textless \ 2.1186) \\ \\ =1-0.98294=\bold{0.0171}[/tex]
Part C:
Find the probability that randomly selected automobile tire has a tread life of less than 38,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, a, is given by:
[tex]P(X \ \textless \ a) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of less than 38,000 miles is given by:
[tex]P(X \ \textless \ 38,000) = P\left(z\ \textless \ \frac{38,000-45,000}{2,360} \right) \\ \\ =P(z\ \textless \ -2.966)=\bold{0.0015}[/tex]
d. Suppose that 6% of all automobile tires with the longest tread life have tread life of at least x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, x, is given by:
[tex]P(X \ \textgreater \ x) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at least x miles is 6%.
Thus:
[tex]P(X \ \textgreater \ x) =0.06 \\ \\ \Rightarrow1 - P(X \ \textless \ x)=0.06 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.06=0.94 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 1.555) \\ \\ \Rightarrow \frac{x-45,000}{2,360}=1.555 \\ \\ \Rightarrow x-45,000=2,360(1.555)=3,669.8 \\ \\ \Rightarrow x=3,669.8+45,000=48,669.8[/tex]
Therefore, the value of x is 48,669.8
e. Suppose that 2% of all automobile tires with the shortest tread life have tread life of at most x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, x, is given by:
[tex]P(X \ \textless \ x) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at most x miles is 2%.
Thus:
[tex]P(X \ \textless \ x)=0.02 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.02=0.98 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 2.054) \\ \\ \Rightarrow \frac{x-45,000}{-2,360}=2.054 \\ \\ \Rightarrow x-45,000=-2,360(2.054)=-4,847.44 \\ \\ \Rightarrow x=-4,847.44+45,000=40,152.56[/tex]
Therefore, the value of x is 40,152.56
Part A:
Find the probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is between two numbers, a and b is given by:
[tex]P(a \ \textless \ X \ \textless \ b) = P(X \ \textless \ b) - P(X \ \textless \ a) \\ \\ P\left(z\ \textless \ \frac{b-\mu}{\sigma} \right)-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles is given by:
[tex]P(42,000 \ \textless \ X \ \textless \ 46,000) = P(X \ \textless \ 46,000) - P(X \ \textless \ 42,000) \\ \\ P\left(z\ \textless \ \frac{46,000-45,000}{2,360} \right)-P\left(z\ \textless \ \frac{42,000-45,000}{2,360} \right) \\ \\ =P(0.4237)-P(-1.271)=0.66412-0.10183=\bold{0.5623}[/tex]
b. Find the probability that randomly selected automobile tire has a tread life of more than 50,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, a, is given by:
[tex]P(X \ \textgreater \ a) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of more than 50,000 miles is given by:
[tex]P(X \ \textgreater \ 50,000) = 1 - P(X \ \textless \ 50,000) \\ \\ =1-P\left(z\ \textless \ \frac{50,000-45,000}{2,360} \right)=1-P(z\ \textless \ 2.1186) \\ \\ =1-0.98294=\bold{0.0171}[/tex]
Part C:
Find the probability that randomly selected automobile tire has a tread life of less than 38,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, a, is given by:
[tex]P(X \ \textless \ a) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of less than 38,000 miles is given by:
[tex]P(X \ \textless \ 38,000) = P\left(z\ \textless \ \frac{38,000-45,000}{2,360} \right) \\ \\ =P(z\ \textless \ -2.966)=\bold{0.0015}[/tex]
d. Suppose that 6% of all automobile tires with the longest tread life have tread life of at least x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, x, is given by:
[tex]P(X \ \textgreater \ x) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at least x miles is 6%.
Thus:
[tex]P(X \ \textgreater \ x) =0.06 \\ \\ \Rightarrow1 - P(X \ \textless \ x)=0.06 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.06=0.94 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 1.555) \\ \\ \Rightarrow \frac{x-45,000}{2,360}=1.555 \\ \\ \Rightarrow x-45,000=2,360(1.555)=3,669.8 \\ \\ \Rightarrow x=3,669.8+45,000=48,669.8[/tex]
Therefore, the value of x is 48,669.8
e. Suppose that 2% of all automobile tires with the shortest tread life have tread life of at most x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, x, is given by:
[tex]P(X \ \textless \ x) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)[/tex]
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at most x miles is 2%.
Thus:
[tex]P(X \ \textless \ x)=0.02 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.02=0.98 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 2.054) \\ \\ \Rightarrow \frac{x-45,000}{-2,360}=2.054 \\ \\ \Rightarrow x-45,000=-2,360(2.054)=-4,847.44 \\ \\ \Rightarrow x=-4,847.44+45,000=40,152.56[/tex]
Therefore, the value of x is 40,152.56
