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2 years ago

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Metro buses are scheduled to arrive at each stop every 30 minutes. If the time a person waits at a bus stop is uniformly distrib

uted and the maximum possible wait time for a bus is 30 minutes, what is the probability that you will wait more than 25 minutes for a bus

1 answer:

Nutka1998 [239]2 years ago

4 0

Answer:

The probability is $P(X > 25 ) = 1.667$

Step-by-step explanation:

From the question we are told that

The maximum wait is $b = 30 \ minutes$

The minimum wait would be $a = 0\ minutes$

Generally the probability that you will wait more than 25 minutes for a bus is mathematically represented as

$P(X > 25 )$

Now given that the time a person waits is uniformly distributed and that the maximum is 30 then we can evaluate the above probability as

$P(X > 25 ) = P(25 < X < 30 ) = \frac{30 - 25}{30 - 0}$

$P(X > 25 ) = 1.667$

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$Data: 5\ 3\ 7\ 8\ 0\ 1\ 0\ 5\ 12\ 10\ 7\ 6\ 7\ 11\ 9$

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$Mean = \frac{\sum x}{n}$

This gives:

$Mean = \frac{5+ 3+ 7+ 8+ 0+ 1+ 0+ 5+ 12+ 10+ 7+ 6+ 7+ 11+ 9}{15}$

$Mean = \frac{91}{15}$

$Mean = 6.07$

Solving (b): The median

Sort the data in ascending order:

$Data: 5\ 3\ 7\ 8\ 0\ 1\ 0\ 5\ 12\ 10\ 7\ 6\ 7\ 11\ 9$

$Sorted: 0\ 0\ 1\ 3\ 5\ 5\ 6\ 7\ 7\ 7\ 8\ 9\ 10\ 11\ 12$

The median is:

$Median = \frac{n + 1}{2}th$

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Solving (c): The mode

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