Question

find the probability using the normal distribution: ( p(-1.72 < z < 1.98) ). use the standard normal distribution table and enter the answer to 4 decimal places.

( p(-1.72 < z < 1.98)=)

Explanation:

Step1: Use the property of normal distribution

We know that \(P(a < z < b)=P(z < b)-P(z < a)\). Here \(a=- 1.72\) and \(b = 1.98\).

Step2: Look up values in the standard - normal table

From the standard - normal distribution table:

• For \(z = 1.98\), \(P(z<1.98)=0.9761\) (by looking up the row \(1.9\) and column \(0.08\) in the standard - normal table).
• For \(z=-1.72\), \(P(z < - 1.72)=1 - P(z < 1.72)\). Looking up \(z = 1.72\) (row \(1.7\) and column \(0.02\)), \(P(z < 1.72)=0.9573\). So \(P(z < - 1.72)=1 - 0.9573=0.0427\).

Step3: Calculate the probability

\(P(-1.72 < z < 1.98)=P(z < 1.98)-P(z < - 1.72)\)
Substitute the values: \(P(-1.72 < z < 1.98)=0.9761-0.0427 = 0.9334\)

Answer:

\(0.9334\)