Question

the most precise value of p(z is greater than -1.61) is _____. select the correct response: 0.946 103 071 85; 0.946 013 071 85; 0.053 698 928 15; 0.053 689 928 15; my answer is not listed among the given choices.; 0.946 301 071 85; 0.053 896 710 15; 0.053 869 928 15

Explanation:

Step1: Recall the property of the standard normal distribution

For a standard normal random variable \( z \), \( P(z > a)=1 - P(z\leq a) \). Also, due to the symmetry of the standard normal curve, \( P(z\leq -a)=1 - P(z\leq a) \) for \( a>0 \), which implies \( P(z > -a)=P(z\leq a) \).

Step2: Use the standard normal table or calculator to find \( P(z\leq 1.61) \)

We know that for \( z = 1.61 \), looking up in the standard normal table (or using a calculator with a normal distribution function), the cumulative probability \( P(z\leq 1.61) \) is approximately \( 0.9463 \).

Since \( P(z > - 1.61)=P(z\leq 1.61) \) (by the symmetry of the standard normal distribution), we have \( P(z > - 1.61)\approx0.9463 \).

Now let's check the given options:

• Option 1: \( 0.94610307185 \)
• Option 2: \( 0.94601307185 \)
• Option 3: \( 0.05369892815 \)
• Option 4: \( 0.05368992815 \)
• Option 5: "My answer is not listed among the given choices."
• Option 6: \( 0.94630107185 \)
• Option 7: \( 0.05389671015 \)
• Option 8: \( 0.05386992815 \)

The value \( 0.94630107185 \) (Option 6) is very close to the calculated value of approximately \( 0.9463 \).

Answer:

0.946 301 071 85 (the sixth option: 0.946 301 071 85)