X follows \mathcal{N}\left(0{,}1\right). The following table is given:
Determine a such that P\left(-a \lt X \lt a\right)=0.4.
We have :
P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)\approx P\left(Z\lt z_{a}\right)-P\left(Z\lt z_{-a}\right)=P\left(Z\lt z_{a}\right)-\left(1-P\left(Z\lt z_{a}\right)\right)=2\cdot P\left(Z\lt z_{a}\right)-1
We are given that:
P\left(-a \lt X \lt a\right)=0.4
So, we can solve the equation:
2\cdot P\left(Z\lt z_{a}\right)-1=0.4
P\left(Z\lt z_{a}\right)=0.7
The table of standard normal probabilities gives the probability of observing an outcome below the given z -score.
Looking at the table of standard normal probabilities, we find that:
z_{a}=0.53
The z -score is the number of standard deviations an observation lies from the mean:
z_{a}=\dfrac{a-\mu}{\sigma}
\mu is the mean and \sigma is the standard deviation.
In our problem, the mean is \mu=0 and the standard deviation is \sigma=1.
So, we can solve the equation:
0.53=\dfrac{a-0}{1}
The solution of the equation is:
a=0.53
a=0.53
Determine a such that P\left(-a \lt X \lt a\right)=0.81.
We have :
P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)\approx P\left(Z\lt z_{a}\right)-P\left(Z\lt z_{-a}\right)=P\left(Z\lt z_{a}\right)-\left(1-P\left(Z\lt z_{a}\right)\right)=2\cdot P\left(Z\lt z_{a}\right)-1
We are given that:
P\left(-a \lt X \lt a\right)=0.81
So, we can solve the equation:
2\cdot P\left(Z\lt z_{a}\right)-1=0.81
P\left(Z\lt z_{a}\right)=0.905
The table of standard normal probabilities gives the probability of observing an outcome below the given z -score.
Looking at the table of standard normal probabilities, we find that:
z_{a}=1.31
The z -score is the number of standard deviations an observation lies from the mean:
z_{a}=\dfrac{a-\mu}{\sigma}
\mu is the mean and \sigma is the standard deviation.
In our problem, the mean is \mu=0 and the standard deviation is \sigma=1.
So, we can solve the equation:
1.31=\dfrac{a-0}{1}
The solution of the equation is:
a=1.31
a=1.31
Determine a such that P\left(-a \lt X \lt a\right)=0.3.
We have :
P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)\approx P\left(Z\lt z_{a}\right)-P\left(Z\lt z_{-a}\right)=P\left(Z\lt z_{a}\right)-\left(1-P\left(Z\lt z_{a}\right)\right)=2\cdot P\left(Z\lt z_{a}\right)-1
We are given that:
P\left(-a \lt X \lt a\right)=0.3
So, we can solve the equation:
2\cdot P\left(Z\lt z_{a}\right)-1=0.3
P\left(Z\lt z_{a}\right)=0.65
The table of standard normal probabilities gives the probability of observing an outcome below the given z -score.
Looking at the table of standard normal probabilities, we find that:
z_{a}=0.39
The z -score is the number of standard deviations an observation lies from the mean:
z_{a}=\dfrac{a-\mu}{\sigma}
\mu is the mean and \sigma is the standard deviation.
In our problem, the mean is \mu=0 and the standard deviation is \sigma=1.
So, we can solve the equation:
0.39=\dfrac{a-0}{1}
The solution of the equation is:
a=0.39
a=0.39
Determine a such that P\left(-a \lt X \lt a\right)=0.64.
We have :
P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)\approx P\left(Z\lt z_{a}\right)-P\left(Z\lt z_{-a}\right)=P\left(Z\lt z_{a}\right)-\left(1-P\left(Z\lt z_{a}\right)\right)=2\cdot P\left(Z\lt z_{a}\right)-1
We are given that:
P\left(-a \lt X \lt a\right)=0.64
So, we can solve the equation:
2\cdot P\left(Z\lt z_{a}\right)-1=0.64
P\left(Z\lt z_{a}\right)=0.82
The table of standard normal probabilities gives the probability of observing an outcome below the given z -score.
Looking at the table of standard normal probabilities, we find that:
z_{a}=0.92
The z -score is the number of standard deviations an observation lies from the mean:
z_{a}=\dfrac{a-\mu}{\sigma}
\mu is the mean and \sigma is the standard deviation.
In our problem, the mean is \mu=0 and the standard deviation is \sigma=1.
So, we can solve the equation:
0.92=\dfrac{a-0}{1}
The solution of the equation is:
a=0.92
a=0.92
Determine a such that P\left(-a \lt X \lt a\right)=0.88.
We have :
P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)\approx P\left(Z\lt z_{a}\right)-P\left(Z\lt z_{-a}\right)=P\left(Z\lt z_{a}\right)-\left(1-P\left(Z\lt z_{a}\right)\right)=2\cdot P\left(Z\lt z_{a}\right)-1
We are given that:
P\left(-a \lt X \lt a\right)=0.88
So, we can solve the equation:
2\cdot P\left(Z\lt z_{a}\right)-1=0.88
P\left(Z\lt z_{a}\right)=0.94
The table of standard normal probabilities gives the probability of observing an outcome below the given z -score.
Looking at the table of standard normal probabilities, we find that:
z_{a}=1.56
The z -score is the number of standard deviations an observation lies from the mean:
z_{a}=\dfrac{a-\mu}{\sigma}
\mu is the mean and \sigma is the standard deviation.
In our problem, the mean is \mu=0 and the standard deviation is \sigma=1.
So, we can solve the equation:
1.56=\dfrac{a-0}{1}
The solution of the equation is:
a=1.56
a=1.56
Determine a such that P\left(-a \lt X \lt a\right)=0.93.
We have :
P\left(-a \lt X \lt a\right)=P\left(X \lt a\right)-P\left(X \lt -a\right)\approx P\left(Z\lt z_{a}\right)-P\left(Z\lt z_{-a}\right)=P\left(Z\lt z_{a}\right)-\left(1-P\left(Z\lt z_{a}\right)\right)=2\cdot P\left(Z\lt z_{a}\right)-1
We are given that:
P\left(-a \lt X \lt a\right)=0.93
So, we can solve the equation:
2\cdot P\left(Z\lt z_{a}\right)-1=0.93
P\left(Z\lt z_{a}\right)=0.965
The table of standard normal probabilities gives the probability of observing an outcome below the given z -score.
Looking at the table of standard normal probabilities, we find that:
z_{a}=1.81
The z -score is the number of standard deviations an observation lies from the mean:
z_{a}=\dfrac{a-\mu}{\sigma}
\mu is the mean and \sigma is the standard deviation.
In our problem, the mean is \mu=0 and the standard deviation is \sigma=1.
So, we can solve the equation:
1.81=\dfrac{a-0}{1}
The solution of the equation is:
a=1.81
a=1.81