Question

Three resistors with resistances R₁, R₂, R₃ are connected in parallel across a battery with voltage V. By Ohm’s law, the current (amps) is

I = V* [ (1/R₁) + (1/R₂) + (1/R₃) ]

Assume that R₁, R₂, R_3, and V are independent random variables

where R₁ ~ Normal (m = 10 ohms, s = 1.5 ohm)

            R₂ ~ Normal (m = 15 ohms, s = 1.5 ohm)

            R₃ ~ Normal (m =20 ohms, s = 1.0 ohms)

            V ~ Normal (m = 120 volts, s = 2.0 volts

(a) Use Monte Carlo Simulation (10,000 random draws from each input random variable) to estimate the mean and standard deviation of the output variable current. (b) Assess whether the output variable current is normally distributed. (c) Assess whether the inverse of current squared (1/ I2 ) is normally distributed. (d) Estimate the probability that the current is less than 25 amps assuming that the inverse of current squared is normally distributed. (e) Compare your answer to (d) with your simulation results – how many of the 10,000 random results for current are below 25 amps via the Stat > Tables > Tally command?

Answer 1

The given formula for current (I) is

I = V* [ (1/R₁) + (1/R₂) + (1/R₃) ]

where R₁, R₂, R₃ are resistance's connected in parallel & V is the battery voltage

    and R₁ ~ Normal (m = 30 ohms, s = 0.60 ohm)

            R₂ ~ Normal (m = 25 ohms, s = 0.60 ohm)

            R₃ ~ Normal (m = 35 ohms, s = 0.60 ohm)

            V ~ Normal (m = 120 volts, s = 1.0 volts)

A) We have to use Monte Carlo Simulation to generate 10,000 random values for current(I).

[The steps are given below to conduct the Monte Carlo simulation by using Minitab software]

Step - 1

First enter the variables name in the worksheet as given in the picture

Step - 2

Generate 10000 observations for R1,R2,R3 and V according to normal distribution with given mean & standard deviation. Use option Calc > Random data >Normal

Then Select the column for R1 and enter the mean & sd, click ok. Repeat the process for R2,R3 and V.

Step - 3

Use option Calc > Calculator to calculate the values for current(I) by using given formula. The formula is entered in the Expression window. Click ok.

we will get results in worksheet like as

Step - 4

To check normality of data generated for current(I), use option Stat > Basic Statistics > Graphical Summary

Then in the summary window, select the variable (I) & confidence interval (95.0) keep as it is and click ok.

The new window will display with all the information that we need.

The summary shows that the p-value for normality test is 0.381 which is greater than significance level (0.05). Therefore we accept the hypothesis that, output variable current is normally distributed. The mean and standard deviation for output variable current is 12.233 and 0.181 respectively.

Step - 5

To check the normality of inverse of current, create a new variable inv(I) in worksheet.

Then use option Calc > Calculator to calculate inverse values of current.

Use graphical summary option as given in step 4 to get the results.

The summary for inverse of current shows that, p-value (0.167) for normality test is greater than significance level (0.05). Therefore we accept the hypothesis that, the inverse of current (1/ I ) is normally distributed.

Step - 6

To find the frequency for current below 12 amps in 10,000 random results, use option tat > Tables > Tally Individual variables

Then select the variable current(I) , tick the cumulative counts. Under session window, we will get the result for frequency for current(I), that is less than 12 amps

D)

956 numbers out of 10000 are below 12 amps. P(current below 12 amps) = 956/10000 = 0.0956 = 0.1 (rounded)

E] We have to estimate the probability that the current is less than 12 amps assuming that the current is normally distributed.

i.e. we have to find P(I<12) assuming I ~ Normal (mean = 12.23, sd = 0.18)

P(I<12) = P(Z < (12-12.23)/0/18)

            = P( Z < - 0.23/0.18)

            = P(Z < - 1.28)

P(I <12) = 0.1 (from Z table)

It seems that, the probability calculated in (D) is same as probability in (E).

Thank you for asking. Please rate my answer...!!