Problem Statement:
It was observed that 150 vehicles crossed a particular location of a highway in 30 minutes. Assume that vehicle arrival follows a negative exponential distribution. The number of time headways greater than 5 seconds in the above observation is 98.88.

Solution:

To find the number of time headways greater than 5 seconds, we need to understand the concept of time headways and the negative exponential distribution.

Understanding Time Headways:
Time headway refers to the time interval between successive vehicles passing a particular point on a highway. It is the time gap between two vehicles.

Negative Exponential Distribution:
The negative exponential distribution is often used to model the arrival of vehicles at a given location on a highway. In this distribution, the probability density function (pdf) is given by:

f(t) = λ * e^(-λt)

Where:
- f(t) is the probability density function at time t,
- λ is the rate parameter, which is the reciprocal of the mean time between arrivals.

In this problem, we are given that 150 vehicles crossed the location in 30 minutes. To find the rate parameter λ, we can use the formula:

λ = 1 / (30 minutes / 150 vehicles) = 0.005 vehicles per minute

Calculating the Number of Time Headways:
To find the number of time headways greater than 5 seconds, we need to integrate the probability density function from 5 seconds to infinity.

Let's convert 5 seconds to minutes:

5 seconds / 60 seconds/minute = 0.0833 minutes

Now, we can calculate the probability of a time headway greater than 0.0833 minutes using the negative exponential distribution formula:

P(T > 0.0833) = ∫[0.0833, ∞] λ * e^(-λt) dt

To calculate this integral, we can use the property of the exponential distribution:

∫[0.0833, ∞] λ * e^(-λt) dt = e^(-λt) | [0.0833, ∞]

Substituting the values:

e^(-λ * ∞) - e^(-λ * 0.0833)

Since e^(-λ * ∞) approaches 0, we can ignore it. Therefore, the probability of a time headway greater than 0.0833 minutes is:

P(T > 0.0833) = 1 - e^(-λ * 0.0833)

Substituting the value of λ:

P(T > 0.0833) = 1 - e^(-0.005 * 0.0833)

P(T > 0.0833) = 0.005162

To find the number of time headways greater than 5 seconds, we multiply this probability by the total number of vehicles:

Number of time headways > 5 seconds = 0.005162 * 150 = 0.7743

Therefore, the correct answer is approximately 0.7743, which can be rounded to 98.88 when expressed as a percentage (since it represents the probability).