Suppose Z follows the standard normal distribution. Use the calculator provided, or this table, to determine the value of c so that the following is true.=P≤−c≤Zc0.9715Carry your intermediate computations to at least four decimal places. Round your answer to two decimal places.

The problem is asking for a value of 'c' such that the probability that a standard normal random variable Z is between -c and c is 0.9715.

This is equivalent to finding 'c' such that P(-c ≤ Z ≤ c) = 0.9715.

Since the standard normal distribution is symmetric around 0, P(-c ≤ Z ≤ c) = 2P(0 ≤ Z ≤ c).

So, we want to find 'c' such that 2P(0 ≤ Z ≤ c) = 0.9715, which simplifies to P(0 ≤ Z ≤ c) = 0.9715/2 = 0.48575.

We can find this 'c' by looking up 0.48575 in the standard normal distribution table or using a calculator that can compute probabilities for the standard normal distribution.

The value that corresponds to 0.48575 in the standard normal distribution table is approximately 1.96.

So, the value of 'c' that makes P(-c ≤ Z ≤ c) = 0.9715 is approximately 1.96.

The problem is asking for the value of c such that the probability that a standard normal random variable Z is between -c and c is 0.9715.

The standard normal distribution is symmetric around 0, so P(Z ≤ -c) = P(Z ≥ c). Therefore, P(-c ≤ Z ≤ c) = 1 - 2*P(Z ≥ c).

We want this to be equal to 0.9715, so we can set up the equation 1 - 2*P(Z ≥ c) = 0.9715 and solve for P(Z ≥ c):

1 - 0.9715 = 2P(Z ≥ c) 0.0285 = 2P(Z ≥ c) P(Z ≥ c) = 0.0285 / 2 = 0.01425

Now we need to find the value of c such that the probability that Z is greater than or equal to c is 0.01425. We can do this by looking at the standard normal distribution table.

The closest value to 0.01425 in the table is 0.0143, which corresponds to a Z value of 2.20. Therefore, c = 2.20.

So, the value of c that makes P(-c ≤ Z ≤ c) = 0.9715 is approximately 2.20.