Question

og (4365 55565 NX! Math 1600 Laboratory Assignment #11 Name Calculating Probabilities for Normal Random Variables Example: The speed of cars on a certain stretch of freeway is normally distributed with mean 65 mph (u - 65) and standard deviation 5 mph (0-5). 1-4142 Find the probability that a car's speed will be X-4 X - 65 (1) greater than 72 mph Let X = car's speed. Then Z- will be a o 5 standard normal random variable. .4142 X - 65 P(X>72) = PT 72 - 65) - P(z>'}) = P(Z > 1.40) 5 72 X -1 - P(Z < 1.40) - 1 -0.9192 -0.0808. 1.46 2 X - 65 55 - 65 .6235 (2) less than 55 mph P(X<55) - PC = P(Z) - P(Z<-2.00 -0.0228 (3) between 61 and 71 mph P( 61 < X <71) 9849-2114 61 - 65 X- 65 -PC = 673 5 5 = P( -0.80 < Z < 1.20 ) = P(Z < 1.20 ) - P(Z < 0.80) .8141 -0.8849 -0.2119 = 0.6730 21 1.2. Use Table 3 to find the following probabilities. For each part, mark the given x-value on the number line on the X-scale, shade the appropriate region under the standard normal curve, convert the x-value to a z-value, mark the given Z-value on the number line on the Z-scale and label the desired probability (the area of the shaded region under the curve). Round calculated 7-values to the nearest 2 decimal places. Problem: The IQ's in a certain population are normally distributed with mean 104 and standard deviation 14. If we select a person at random from this population, find the probability that their IQ will be 71 - <?=65 65 -0.30 (1) less than 100 1oy 코 (2) greater than 140 104 o X dhe (3) between 95 and 115 104 0 Z

Answer 1

1) 0.3874

2)0.0051

3) 0.5229

Problem- Mean, u = 104 Standard deviations=14 0.3874 (1) P(x< 100) 2 = X-4 14 = -0.280 loo-loy 100 104 Plx<100)=P(22-0-206) = 0.3874 P(x-140) Z=140-lon E 19 = 2.57 0.005 1 E PCX >140) = P(732.59) 140 4648 204611 soil - 0.0051 (3). doo 0.5२२१ 721 p/95 eX=115) = p (15kg cocaine = ploche 2 <0.786) = P(20 786) - P/26 - 0.84) = 0.7840-0.2611 0.5739 Is 104 lis