P(-0.91≤Z≤c)=0.7930

Question

Suppose follows the standard normal distribution. Use the calculator provided, or this table, to determine the value of 

 so that the following is true.

P(-0.91≤Z≤c)=0.7930

Answer 1

find the value of "c" such that P(-0.91 ≤ Z ≤ c) = 0.7930, we need to use a standard normal distribution table or a calculator with a standard normal distribution function.

The standard normal distribution table provides the cumulative probability for values of the standard normal random variable "Z" (mean = 0, standard deviation = 1). In this case, we are given the cumulative probability of 0.7930 for the range -0.91 to "c."

Using the standard normal distribution table or a calculator, we can look up the corresponding Z-score for the cumulative probability of 0.7930. The Z-score will be the value of "c" that we are looking for.

For the Z-score of 0.7930, the corresponding "c" value is approximately 0.8531.

So, P(-0.91 ≤ Z ≤ 0.8531) ≈ 0.7930.