Questions: Question 5
0 / 2 pts
3 ↔ 99
Details
The arrows can only be dragged to z scores that are accurate to 1 place after the decimal point (these values correspond to the tick marks on the horizontal axis). Select from the drop down menu to shade to the left, to the right, between or left and right of the z-score. You may round probabilities to three decimal places.
a) Sketch the region corresponding to the statement P(-0.9<Z<1.5)
Shade: Left of a value v̂. Click and drag the arrows to adjust the values.
b) Find P(-0.9<Z<1.5)
Transcript text: Question 5
$0 / 2$ pts
$3 \rightleftarrows 99$
Details
The arrows can only be dragged to $z$ scores that are accurate to 1 place after the decimal point (these values correspond to the tick marks on the horizontal axis). Select from the drop down menu to shade to the left, to the right, between or left and right of the z-score. You may round probabilities to three decimal places.
a) Sketch the region corresponding to the statement $P(-0.9
Solution
Solution Steps
Step 1: Identify the Z-scores
The problem requires us to find the probability for the range \(-0.9 < Z < 1.5\). The Z-scores are \(-0.9\) and \(1.5\).
Step 2: Use the Standard Normal Distribution Table
Look up the cumulative probabilities for the Z-scores in the standard normal distribution table:
For \(Z = -0.9\), the cumulative probability is approximately \(0.1841\).
For \(Z = 1.5\), the cumulative probability is approximately \(0.9332\).
Step 3: Calculate the Probability
To find the probability that \(Z\) is between \(-0.9\) and \(1.5\), subtract the cumulative probability at \(-0.9\) from the cumulative probability at \(1.5\):
\[ P(-0.9 < Z < 1.5) = P(Z < 1.5) - P(Z < -0.9) \]
\[ P(-0.9 < Z < 1.5) = 0.9332 - 0.1841 \]
\[ P(-0.9 < Z < 1.5) = 0.7491 \]