Assume the random variable X is normally distributed with mean mu?equals=5050 and standard deviation sigma?equals=77. Compute the probability. Be sure to draw a normal curve with the area corresponding to the probability shaded. ?P(5757less than or equals?Xless than or equals?6868?) Which of the following normal curves corresponds to ?P(5757less than or equals?Xless than or equals?6868?)? A. 575750506868 A normal curve has a horizontal axis with three labeled coordinates, 50, 57, and 68. The curve's peak is near the top of the graph at horizontal coordinate 50. Three vertical line segments run from the horizontal axis to the curve at 50, 57, and 68. The area under the curve to the left of the vertical line segment at 57 is shaded. B. 575750506868 A normal curve has a horizontal axis with three labeled coordinates, 50, 57, and 68. The curve's peak is near the top of the graph at horizontal coordinate 50. Three vertical line segments run from the horizontal axis to the curve at 50, 57, and 68. The area under the curve between the vertical line segments at 57 and 68 is shaded. C. 575750506868 A normal curve has a horizontal axis with three labeled coordinates, 50, 57, and 68. The curve's peak is near the top of the graph at horizontal coordinate 50. Three vertical line segments run from the horizontal axis to the curve at 50, 57, and 68. The area under the curve between the vertical line segments at 50 and 57 is shaded. ?P(5757less than or equals?Xless than or equals?6868?)equals=nothing ?(Round to four decimal places as? needed.)
Population: m = 50 and standard deviation s = 7. P(x > 65) = 1- P(z?15/7) = 1- P(z ?2.1429) = 1 -.9839 = .016 0r 1.16% P(x > 65): shade area to the right of z = 2.1429 for 28. P(z ? 2.5714) - P(z ? .8571) find z's and shade area between them For the normal distribution: Below: z = 0, z = ± 1, z= ±2 , z= ±3 are plotted. Area under the standard normal curve to the left of the particular z is P(z) Note: z = 0 (x value: the mean) 50% of the area under the curve is to the left and 50% to the right.
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