Question

1) Rate data often follow a lognormal distribution. Average power usage (dB per hour) for a particular company is studied and is known to have a lognormal distribution with parameters μ = 4 and σ = 2. What is the probability that the company uses more than 270 dB during any particular hour?

2) A certain type of device has an advertised failure rate of 0.01 per hour. The failure rate is constant and the exponential distribution applies. (a) What is the mean time to failure? (b) What is the probability that 200 hours will pass before a failure is observed?

3) The elongation of a steel bar under a particular load has been established to be normally distributed with a mean of 0.05 inch and σ = 0.01 inch. Find the probability that the elongation is (a) above 0.1 inch; (b) below 0.04 inch; (c) between 0.025 and 0.065 inch.

4) A controlled satellite is known to have an error (distance from target) that is normally distributed with mean zero and standard deviation 4 feet. The manufacturer of the satellite defines a success as a firing in which the satellite comes within 10 feet of the target. Compute the probability that the satellite fails.

Answer 1

1. SOLUTION:

Let the random variable X be a power usage of a company.

Determine the probability that the company uses more than 270 dB during any particular hour.

From the given information, the random variable in(X) has a normal distribution with mean
_w=4
and standard deviation
0 = 2

Now, compute the required probability as follows:

P(X > 270) = P(in(x) > In(270))

In(X) - In(270) - 4 = P

= P(Z < -0.80)

= NORM SDIST(-0.80)

{Use Excel function or Z-table}

=0.2119

As a result, the probability that the company uses more than 270 dB during particular hour is 0.2119.

2.SOLUTION:

Given that the life of a certain type of device has an advertised failure rate of
_=0.01.

And the failure rate follows exponential distribution.

(a)

The aim is to find the mean time to failure.

The failure rate
_=0.01.

The mean time to failure , when an exponential distribution applies,

001 = 100-Yea

Mean of the failure time is 100 hours.

(b)

The aim is to find the probability that 200 hours will pass before a failure is observed.

P(X > 200) = 1- P1 < 200)

1-11--0.01x2007 = 1 - [1 – 0.135335

= 0.135335.

Hence, the probability that 200 hours will pass before a failure is 0.1353.

3.SOLUTION:

a)

Now , we want to find the probability that the elongation is above 0.1 inch

we find that
_{1 -}

10.1 -0.05 10.01 - -

Hence, the required probability is,

P(X>0.1) = 1- P(X<0.1)

1-P(Z <5) = 1-1=0.

b)

Now, we want to find the probability that the elongation is below 0.04 inch

we find that
_{1 -}

0.04 -0.05 2 = -1. 0.01

Hence, the required probability is,

P(X <0.04) = P(Z < -1) = 0.1587.

c)

Now , we want to find the probability that the elongation is between 0.025 and 0.065 inch.

We find that

1 -μ 21 =- 0.025 – 0.05 ο = -2.5. 0.01

1 -μ 22 =- 0.065 – 0.05 0.01 - = 1.5.

Hence, the required probability is,

P(0.025 < X < 0.065) = P(-2.5 <<1.5)

Plz<1.5) - P(Z < -2,5) = 0.9332 -0.0062 = 0.927.

4.SOLUTION:

Let the random variable X represent the distance from target.

From the given information the random variable X follows normal distribution with mean
_{0 = 11}
and the standard deviation
0 = 4.

The probability that the satellite comes within 10 feet the target is,

10 - 0 P(X < 10) = P.

( From standard normal tables)

= P(Z < 10) = P(Z < 2.5) = 0.9938.

The probabilities that satellite fails is,

P( Satellite fails) = 1-P ( X _$\leq$ 10 )

= 1 = r( - - 10 - 0)

= 1- P(Z < 2.5) = 0.0062.

Hence , the required probability is 0.0062.

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