Answer :
To find the approximate value of [tex]\( T \)[/tex] for the interval [tex]\( T \leq \Theta \leq 1.10 \)[/tex] in a standard normal distribution, we first need to identify the corresponding z-scores and their probabilities.
Using the standard normal table provided, here are the probabilities for given [tex]\( z \)[/tex]-scores:
[tex]\[ \begin{array}{|c|c|} \hline z & \text{Probability} \\ \hline 0.00 & 0.5000 \\ \hline 0.16 & 0.5636 \\ \hline 0.22 & 0.5871 \\ \hline 0.78 & 0.7823 \\ \hline 1.00 & 0.8413 \\ \hline 1.16 & 0.8770 \\ \hline 1.78 & 0.9625 \\ \hline 2.00 & 0.9772 \\ \hline \end{array} \][/tex]
1. Find the probability corresponding to [tex]\( \Theta = 1.10 \)[/tex]. We do not have a direct value for [tex]\( z = 1.10 \)[/tex] from the table, but we can interpolate. [tex]\( z = 1.10 \)[/tex] lies between [tex]\( z = 1.00 \)[/tex] and [tex]\( z = 1.16 \)[/tex]. The corresponding probabilities are:
- For [tex]\( z = 1.00 \)[/tex], [tex]\( P(Z \leq 1.00) = 0.8413 \)[/tex]
- For [tex]\( z = 1.16 \)[/tex], [tex]\( P(Z \leq 1.16) = 0.8770 \)[/tex]
We can estimate the probability for [tex]\( z = 1.10 \)[/tex] using linear interpolation.
[tex]\[ \text{Interpolated probability} = 0.8413 + \left( \frac{1.10 - 1.00}{1.16 - 1.00} \right) (0.8770 - 0.8413) \][/tex]
Simplifying,
[tex]\[ \text{Interpolated probability} = 0.8413 + \left( \frac{0.10}{0.16} \right) \times 0.0357 \][/tex]
[tex]\[ = 0.8413 + 0.223 \times 0.0357 \][/tex]
[tex]\[ = 0.8413 + 0.0223 \][/tex]
[tex]\[ \approx 0.8636 \][/tex]
2. We aim to find [tex]\( T \)[/tex] such that the total probability interval [tex]\( P(T \leq \Theta \leq 1.10) \)[/tex] is equal to one of the given choices (22%, 66%, 78%, or 88%).
The total probability over the range [tex]\( T \leq \Theta \leq 1.10 \)[/tex]:
[tex]\[ P(T \leq \Theta \leq 1.10) = P(\Theta \leq 1.10) - P(\Theta \leq T) \][/tex]
With [tex]\( P(\Theta \leq 1.10) \approx 0.8636 \)[/tex]:
Let's check these given percentages one by one:
- For 22%:
[tex]\[ 0.22 = 0.8636 - P(\Theta \leq T) \][/tex]
[tex]\[ P(\Theta \leq T) \approx 0.8636 - 0.22 = 0.6436 \][/tex]
We need to find the [tex]\( z \)[/tex]-score for [tex]\( P(\Theta \leq T) \approx 0.6436 \)[/tex]. Checking table values, [tex]\( z \approx 0.36 \)[/tex].
- For other percentages, corresponding intervals and calculations suggest higher probabilities, requiring corresponding [tex]\( z \)[/tex]-scores greater than 0.
Since we seek exact nearest value meeting one provided, matching within closest given result, nearest within this calculation to [tex]\( z \)[/tex]-frame, interval approximately aligning to tabulated value derived corresponds closest to 66%.
Using the standard normal table provided, here are the probabilities for given [tex]\( z \)[/tex]-scores:
[tex]\[ \begin{array}{|c|c|} \hline z & \text{Probability} \\ \hline 0.00 & 0.5000 \\ \hline 0.16 & 0.5636 \\ \hline 0.22 & 0.5871 \\ \hline 0.78 & 0.7823 \\ \hline 1.00 & 0.8413 \\ \hline 1.16 & 0.8770 \\ \hline 1.78 & 0.9625 \\ \hline 2.00 & 0.9772 \\ \hline \end{array} \][/tex]
1. Find the probability corresponding to [tex]\( \Theta = 1.10 \)[/tex]. We do not have a direct value for [tex]\( z = 1.10 \)[/tex] from the table, but we can interpolate. [tex]\( z = 1.10 \)[/tex] lies between [tex]\( z = 1.00 \)[/tex] and [tex]\( z = 1.16 \)[/tex]. The corresponding probabilities are:
- For [tex]\( z = 1.00 \)[/tex], [tex]\( P(Z \leq 1.00) = 0.8413 \)[/tex]
- For [tex]\( z = 1.16 \)[/tex], [tex]\( P(Z \leq 1.16) = 0.8770 \)[/tex]
We can estimate the probability for [tex]\( z = 1.10 \)[/tex] using linear interpolation.
[tex]\[ \text{Interpolated probability} = 0.8413 + \left( \frac{1.10 - 1.00}{1.16 - 1.00} \right) (0.8770 - 0.8413) \][/tex]
Simplifying,
[tex]\[ \text{Interpolated probability} = 0.8413 + \left( \frac{0.10}{0.16} \right) \times 0.0357 \][/tex]
[tex]\[ = 0.8413 + 0.223 \times 0.0357 \][/tex]
[tex]\[ = 0.8413 + 0.0223 \][/tex]
[tex]\[ \approx 0.8636 \][/tex]
2. We aim to find [tex]\( T \)[/tex] such that the total probability interval [tex]\( P(T \leq \Theta \leq 1.10) \)[/tex] is equal to one of the given choices (22%, 66%, 78%, or 88%).
The total probability over the range [tex]\( T \leq \Theta \leq 1.10 \)[/tex]:
[tex]\[ P(T \leq \Theta \leq 1.10) = P(\Theta \leq 1.10) - P(\Theta \leq T) \][/tex]
With [tex]\( P(\Theta \leq 1.10) \approx 0.8636 \)[/tex]:
Let's check these given percentages one by one:
- For 22%:
[tex]\[ 0.22 = 0.8636 - P(\Theta \leq T) \][/tex]
[tex]\[ P(\Theta \leq T) \approx 0.8636 - 0.22 = 0.6436 \][/tex]
We need to find the [tex]\( z \)[/tex]-score for [tex]\( P(\Theta \leq T) \approx 0.6436 \)[/tex]. Checking table values, [tex]\( z \approx 0.36 \)[/tex].
- For other percentages, corresponding intervals and calculations suggest higher probabilities, requiring corresponding [tex]\( z \)[/tex]-scores greater than 0.
Since we seek exact nearest value meeting one provided, matching within closest given result, nearest within this calculation to [tex]\( z \)[/tex]-frame, interval approximately aligning to tabulated value derived corresponds closest to 66%.
