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for a standard normal distribution, find the approximate value of $p(z …

Question

for a standard normal distribution, find the approximate value of $p(z \geq -1.25)$. use the portion of the standard normal table below to help answer the question.

zprobability
0.250.5987
1.000.8413
1.250.8944
1.500.9332
1.750.9599

\bigcirc 11%\bigcirc 39%\bigcirc 61%\bigcirc 89%

Explanation:

Step1: Recall Normal Distribution Symmetry

The standard normal distribution is symmetric about \( z = 0 \). So, \( P(z \geq -a)=1 - P(z < -a)=1 - [1 - P(z < a)] = P(z < a) \) (using symmetry \( P(z < -a)=1 - P(z < a) \)). For \( a = 1.25 \), we use \( P(z \geq -1.25)=P(z < 1.25) \) (since \( z \) is continuous, \( P(z = -1.25)=0 \)).

Step2: Find \( P(z < 1.25) \) from Table

From the given standard normal table, when \( z = 1.25 \), the probability \( P(z < 1.25) = 0.8944 \). To convert this to a percentage, multiply by 100: \( 0.8944\times100\approx89\% \).

Answer:

89% (corresponding to the option with 89%)