# Probability: Standard Deviation and Pic

**Topics:**Standard deviation, Probability theory, Arithmetic mean

**Pages:**3 (558 words)

**Published:**June 18, 2012

n = 100

µ = 7µ[pic] = 7

σ = 2.41σ[pic] = 2.41 /[pic]

|X |2 |3 |4 |5 | |P(X=x) |0.3333 |0.3333 |0.3333 | 1.00 | |P.X |199.9800 |99.9900 |0.0000 | 299.97 | |P².X | 119,988.00 | 29,997.00 | - | 149,985.00 |

σ²= 149,985.00 - 299.97²

= 60,003

σ= [pic]

= 244.9551

P([pic]≥320)= 1 - P ( Z < [pic] - µ[pic] )

σ[pic]

= 1 - P ( Z < 320-299.97 )

244.9551/[pic]

= 1 - P ( Z < 20.03 )

14.1425

= 1 - P ( Z < 1.42)

= 1 - 0.9222

= 0.0778 is the probability that at least 320 will attend.

1 - NORMDIST(320,299.97,244.9551/SQRT(300),1) = 0.0783

(c) A circuit contains three resistors wired in series. Each is rated at 6 ohms. Suppose, however, that the true resistance of each one is a normally distributed random variable with a mean of 6 ohms and a standard deviation of 0.3 ohm. What is the probability that the combined resistance will exceed 19 ohms? How "precise" would the manufacturing process have to be to make the probability less than 0.005 that the combined resistance of the circuit would exceed 19 ohms?

n = 3

µ = 6µ[pic] = 6

σ = 0.3σ[pic] = 0.3 /[pic]

[pic]= 19 / 3

P ([pic]≥ 6.33)= 1- P ( Z < [pic] - µ[pic] )

σ[pic]

= 1 - P (Z < 6.33-6 )

0.3 /[pic]

= 1...

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