Why in question#2 (2.2f) the way of solving the exercise is
different from the previous question?
I mean, I do not get why we use F(2) -
F(1), why not F(2) -
R(1) ?
Since both questions are almost the same (it seems), should not
they be solve using the same way?
I really do not when to use F and when to use R.
Why In Question 2 2 2f The Way Of Solving The Exercise Is Different From The Previous Question I Mean I Do Not Get W 1 (544.62 KiB) Viewed 89 times
Z 65 65-72 10 = -0.70 The final scores on a test are normally distributed with a mean of 72 and a standard deviation of 10. What is most nearly the probability that a student's score will be between 65 and 78? 78-72 Zg =+0.6 10 78 65 T A. 0.4196 B. 0.4837 C. 0.5161 D. 0.6455 F(0.60) - 0.7257 R (0.70)= 0.2420 area - 0.7257-02420 = 0.4837 Problem 2.2 f) The average useful life of a typical dishwasher is 10 years with standard deviation of 2 years. Calculate the probability of this dishwasher lasting between 12 to 14 years. Assume that useful life of dishwasher follows normal distribution. According to the problem statement X1 = 12 x2 = 14 u = 10,0 = 2. Since, u = 0 &01 this distribution should be standardized as shown below: 12 – 10 Z1 = = 2 = 1 2 NU 2 22 14 – 10 4 =-= 2 2 2 21 SZ < 22 P(Z <Z SZ2) = F(2) – F(1) = 0.9772 – 0.8413 = 0.1359
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