Question

For the standard normal distribution shown on the​ right, find the probability of z occurring in the indicated region. 0.61 -0.93 z A normal curve is over a horizontal z-axis. Vertical line segments extend from the horizontal axis to the curve at negative 0.93 and 0.61. The area under the curve between negative 0.93 and 0.61 is shaded. The probability is nothing. ​(Round to four decimal places as​ needed.)

Answer 1

Solution: We need to find P(-0.93 < z < 0.61) P(-0.93 z 0.61) = P(z < 0.61) - P(z -0.93) P(-0.93 z 0.61) = P(z 0.61) - P(z -0.93) From normal distribution table; P(z 0.61) - P(z -0.93) 0.7291- 0.1762 Hence, P(-0.93 z 0.61) 0.5529 0.5529