Question

The transmission delay between two linked wireless devices is a normal variable with a mean of 60 milliseconds and a standard deviation of 5 milliseconds.

a. What is the probability that a transmission delay is more than 65 milliseconds?

b. What is the probability that a transmission delay is between 55 and 70 milliseconds?

c. What is the probability that a transmission delay is more than 45 milliseconds?

d. Agents for the NSA notice a transmission delay greater than 80 milliseconds. Is this a rare enough event to warrant suspicion that enemy agents are dampening the signal strength?

Answer 1

Given = 60, = 5

To find the probability, we need to find the z scores.

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(a) For P (X > 65) = 1 - P (X < 65), as the normal tables give us the left tailed probability only.

For P( X < 65)

Z = (65 – 60)/5 = 1

The probability for P(X < 1) from the normal distribution tables is = 0.813

Therefore the required probability = 1 – 0.8413 = 0.1587

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(b) Between 55 and 70 = P(55 < X < 70) = P(X < 70) - P(X < 55)

For P(X < 70) ; z = (70 - 60) / 5 = 2. The p value at this score is = 0.9772

For P(X < 55) ; z = (55 - 60) / 5 = -1. The p value at this score is = 0.1587

Therefore the required probability is 0.9772 – 0.1587 = 0.8185

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(c) For P (X > 45) = 1 - P (X < 45), as the normal tables give us the left tailed probability only.

For P( X < 45)

Z = (45 – 60)/5 = -3

The probability for P(X < 45) from the normal distribution tables is = 0.0013

Therefore the required probability = 1 – 0.0013 = 0.9987

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(d) For P (X > 80) = 1 - P (X < 80), as the normal tables give us the left tailed probability only.

For P( X < 80)

Z = (80 – 60)/5 = 4

The probability for P(X < 80) from the normal distribution tables is = 1

Therefore the required probability = 1 – 1 = 0

Yes, this is a rare enough event as the probability is less than 0.05 (5%) or more than 3 standard deviations.

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