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Statistics / Elementary Statistics: A Step By Step Approach 9 / Chapter 6 / Problem 24CQ

Elementary Statistics: A Step By Step Approach | 9th Edition | ISBN: 9780073534985 | Authors: Allan Bluman

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Passengers on a BusOn the daily run of an express bus, the

Chapter 6, Problem 24CQ

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QUESTION:

On the daily run of an express bus, the average number of passengers is 48. The standard deviation is 3. Assume the variable is normally distributed. Find the probability that the bus will have

a. Between 36 and 40 passengers

b. Fewer than 42 passengers

c. More than 48 passengers

d. Between 43 and 47 passengers

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QUESTION:

On the daily run of an express bus, the average number of passengers is 48. The standard deviation is 3. Assume the variable is normally distributed. Find the probability that the bus will have

a. Between 36 and 40 passengers

b. Fewer than 42 passengers

c. More than 48 passengers

d. Between 43 and 47 passengers

ANSWER:

Answer

Step 1 of 4

(a)

We are asked to find the probability that the bus will have between 36 and 40 passengers.

We have given an average number of passengers is 48 and the standard deviation is 3.

Let $X$ denote the number of passengers in the bus.

We need to find $P(36\leq{}X\leq{}40).$

The $z-value$ corresponding to $40$ is,

$z=\frac{X-\mu{}}{\sigma{}}=\frac{40-48}{3}=-2.67$

The $z-value$ corresponding to $36$ is,

$z=\frac{X-\mu{}}{\sigma{}}=\frac{36-48}{3}=-4$

Hence we can write,

$P(36\leq{}X\leq{}40)=P(Z\leq{}-2.67)-P(Z\leq{}-4)$

From the standard normal table, we can find the value of probability,

$P(Z\leq{}-2.67)=0.038$

$P(Z\leq{}-4)=0$

$P(36\leq{}X\leq{}40)=P(Z\leq{}-2.67)-P(Z\leq{}-4)=0.038-0=0.038$

Hence the probability that the bus will have between 36 and 40 passengers is $0.038.$

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