Question

for a standard normal distribution, find the approximate value of ( p(-0.78 leq z leq 1.16) ). use the portion of the standard normal table below to help answer the question.

( z )	probability
0.16	0.5636
0.72	0.5871
0.78	0.7823
1.00	0.8413
1.16	0.8770
1.78	0.9625
2.00	0.9772

• 22%
• 66%
• 78%
• 88%

Explanation:

Step1: Recall the property of standard normal distribution

For a standard normal distribution, \( P(-a \leq z \leq b)=P(z \leq b)-P(z \leq -a) \), and due to symmetry \( P(z \leq -a) = 1 - P(z \leq a) \), so \( P(-a \leq z \leq b)=P(z \leq b)+P(z \leq a)-1 \) when \( a>0 \). Here \( a = 0.78 \) and \( b = 1.16 \).

Step2: Find \( P(z \leq 1.16) \) and \( P(z \leq 0.78) \) from the table

From the given standard normal table, \( P(z \leq 1.16)=0.8770 \) and \( P(z \leq 0.78) = 0.7823 \).

Step3: Calculate \( P(-0.78 \leq z \leq 1.16) \)

Using the formula \( P(-0.78 \leq z \leq 1.16)=P(z \leq 1.16)+P(z \leq 0.78)-1 \)
Substitute the values: \( 0.8770 + 0.7823-1=0.6593\approx 66\% \)

Answer:

66% (corresponding to the option with 66%)