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Voltage sags and swells. Refer to the Electrical Engineering (Vol. 95, 2013) study of the power quality of a transformer, Exercise 2.127 (p. 132). Recall that two causes of poor power quality are 鈥渟ags鈥� and 鈥渟wells.鈥�. (A sag is an unusual dip, and a swell is an unusual increase in the voltage level of a transformer.) For Turkish transformers built for heavy industry, the mean number of sags per week was 353, and the mean number of swells per week was 184. As in Exercise 2.127, assume the standard deviation of the sag distribution is 30 sags per week, and the standard deviation of the swell distribution is 25 swells per week. Also, assume that the number of sags and number of swells is both normally distributed. Suppose one of the transformers is randomly selected and found to have 400 sags and 100 swells in a week.

a. What is the probability that the number of sags per week is less than 400?

b. What is the probability that the number of swells per week is greater than 100?

Step by step solution

01

Given information

Referring to the Electrical Engineering (Vol. 95, 2013) study power quality of a transformer, exercise 2.127, the distribution of the number of sags per week follows a normal distribution with a mean of 353 and a standard deviation of 30. The distribution of forecast errors from sell-side analysts follows a normal distribution with a mean of 184 and a standard deviation of 25.

02

Step 2:(a) Calculate the probability

$x鈥�鈥�~鈥�鈥�N\left(渭,{蟽}^2\right)$where$渭=353鈥�鈥�鈥�and鈥�鈥�鈥�蟽=30$

$x=400$

The z-score is,

$\begin{array}{rcl}z& =& \frac{x-渭}{蟽}\\ & =& \frac{400-353}{30}\\ & =& 1.566667\\ & 鈮�& 1.57\\ & & \end{array}$

$\begin{array}{rcl}P\left(x<400\right)& =& P\left(z<1.57\right)\\ & =& 0.9417924\\ & 鈮�& 0.9418\\ & & \end{array}$

$P\left(x<400\right)=0.9418$

Therefore, the probability that the number of sags per weekis less than 400 is0.9418.

03

Step 3:(b) Calculate the probability

$x鈥�鈥�~鈥�鈥�N\left(渭,{蟽}^2\right)$where$渭=184鈥�鈥�鈥�and鈥�鈥�鈥�蟽=25$

$x=100$

The z-score is,

$\begin{array}{rcl}z& =& \frac{x-渭}{蟽}\\ & =& \frac{100-184}{25}\\ & =& -3.36\\ & & \end{array}$

$\begin{array}{rcl}P\left(x>100\right)& =& 1-P\left(z<-3.36\right)\\ & =& 1-\left(1-P\left(z<3.36\right)\right)\\ & =& 1-1+0.9996103\\ & =& 0.9996103\\ & 鈮�& 0.9996\\ & & \end{array}$

$P\left(x>100\right)=0.9996$

Therefore, the probability that the number of swells per week isgreater than 100 is0.9996.

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