Final answer:
The variance for a normal distribution where the probability of falling within a circle of radius 1 is 0.75 is assumed to be 1, as this radius corresponds to 1 standard deviation in a standard normal distribution which encompasses approximately 75% of the probability.
Step-by-step explanation:
The question revolves around finding the variance of a normally distributed random variable given that the probability of this variable falling within a circle of radius 1 is 0.75. This is a problem that involves understanding the properties of the normal distribution and how to calculate probabilities using the z-score and variance.
To begin with, a normal distribution is symmetric about the mean, and the probability of falling within a certain range can be determined by the z-score. You can use a calculator (like the TI-83, 83+, or 84+ series) to find the z-score that corresponds to the cumulative probability up to the edge of the circle.
The command invNorm(0.875,0,1) could be used to find the z-score that leaves 0.125 (which is 0.25/2 for both tails) in the upper tail. However, this does not immediately give us the variance.
To find the variance given that the probability is 0.75 within the circle, we must know the standard deviation. Since the question does not provide this, we have to assume that the standard deviation is such that the radius of the circle corresponds to a z-score that encapsulates 75% of the distribution.
If we assume the circle is centered at the mean, then we need to find the z-score that captures 37.5% of the distribution above and below the mean (since the circle is symmetrical). This z-score can be found using invNorm(0.375). We know from the standard normal distribution that a z-score circa 0.6745 captures approximately 25% of the distribution one side of the mean, adding up to 50% for both sides.
Therefore, a z-score of approximately 1 would enclose around 75% of the values in total, meaning the radius 1 corresponds to 1 standard deviation, implying that the variance (²) would be 1² = 1.