Question

Suppose the current measurements in a strip of wire are assumed to follow a normal distribution with a mean of 10 mA and a variance of 4 (mA)^2.

a. What is the probability that a measurement will exceed 13 mA?

b. What is the probability that a current measurement is between 9 and 11 mA?

c. Determine the value for which the probability that a current measurement is below this value is 0.98.

Please show all steps and explanations. Thank you

Answer 1

Given,

= 10 , = sqrt(4) = 2

We convert this to standard normal as

P(X < x) = P(Z < ( x - ) / )

a)

P(X > 13) = P(Z > ( 13 - 10) / 2)

= P(Z > 1.5)

= 0.0668 (From Z table)

b)

P(9 < X < 11) = P(X < 11) - P(X < 9)

= P(Z < (11 - 10) / 2) - P(Z < ( 9 - 11) / 2)

= P(Z < 1) - P(Z < -1)

= 0.8413 - 0.1587  (From Z table)

= 0.6826

c)

We have to calculate x such that

P(X < x) = 0.98

P(Z < (x -   ) / ) = 0.98

From Z table z-score for the probability of 0.98 is 2.0537

(x -   ) / = 2.0537

( x - 10) / 2 = 2.0537

x = 14.1074