Question

(B) An airline knows from experience that the distribution of the number of suitcases that get lost each week on a certain route is normal with μ (mean) = 15.5 and σ (standard deviation) = 3.6. What is the probability that during a given week the airline will lose between 11 and 19 suitcases?

Answer 1

Answer:

The correct answer is "0.7289".

Step-by-step explanation:

The given values are:

Mean,

$\mu = 15.5$

Standard deviation,

$\sigma = 3.6$

As we know,

⇒ $z = \frac{(x - \mu)}{\sigma}$

The probability will be:

⇒ $P(11< x< 19) = P(\frac{11-15.5}{3.6}$

                             $=P(z< 0.9722)-P(z< -1.25)$

By using the z table, we get

                             $=0.8345-0.1056$

                             $=0.7289$