Question

In each problem, make sure that you are clearly defining random variables, stating their distributions, and writing down the formulas that you are using. (That is, write down the pmf, write down mean and variance formulas.)

One of the most common places where normal distributions naturally arise is from the distribution of measurement error. A good example of this would be the old fashioned resistors which are used in electrical circuits. Factories can’t produce perfectly accurate products, so these resistors always display two numbers – an advertised resistance as well as what is called a tolerance. Suppose we have a resistor whose colors indicate a resistance value of 10 kOhms with a tolerance of 5%. The way we are meant to interpret these two numbers is that 10 kOhms is the mean of a normal distribution for the random variable X = the resistance of a single resistor produced at the factory, with a standard deviation which is related to the 5% tolerance. For the purposes of this problem we’ll assume that this means the standard deviation is 0.05. (This number probably actually corresponds to several standard deviations, but who cares.) Calculate the following probabilities about X. Don’t use the empirical rule here. It is highly encouraged that you find these by taking z-scores and using a table, just to confirm that you know how to do it and understand the mechanics.

a. P(10 ≤ X ≤ 10.05)

b. P(9.92 ≤ X ≤ 10.02)

c. P(X ≤ 10.037)

d. P(9.97 ≤ X)

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	10
std deviation   =σ=	0.0500

probability =

P(10<X<10.05)

=

P(0<Z<1)=

0.8413-0.5=

0.3413

b)

probability =

P(9.92<X<10.02)

=

P(-1.6<Z<0.4)=

0.6554-0.0548=

0.6006

c)

probability =

P(X<10.037)

=

P(Z<0.74)=

0.7704

d)

probability =

P(X>9.97)

=

P(Z>-0.6)=

1-P(Z<-0.6)=

1-0.2743=

0.7257