Final answer:
To find the probability that Z falls between -1.90 and 0, subtract the area to the left of Z = -1.90 from the area to the left of Z = 0, which gives P(-1.90 ≤ Z ≤ 0) = 0.4710 after rounding.
Step-by-step explanation:
The question asks us to find the probability that a standard normal variable Z falls between -1.90 and 0. To solve this, we can use a Z-table, which provides the area (probability) to the left of a given Z-score.
To find P(-1.90 ≤ Z ≤ 0), we first look up the probability to the left of Z = -1.90, which is typically around 0.0287. Since Z = 0 corresponds to the mean of the standard normal distribution, the area to the left is 0.5. To find the probability that Z falls between -1.90 and 0, we subtract the area to the left of Z = -1.90 from the area to the left of Z = 0. Therefore:
P(-1.90 ≤ Z ≤ 0) = P(Z < 0) - P(Z < -1.90)
= 0.5 - 0.0287
= 0.4713
So the correct answer is c) 0.4710, after rounding to four decimal places.