Question

A local concert center reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows. Round the standard deviation to three decimal places to work the problem.
A) 0.7967
B) 0.3187
C) 0.2451
D) 0.754

Answer 1

Answer: C

Step-by-step explanation:

Answer 2

The probability that there will be fewer than 20 no-shows will be "0.2451".

According to the question,

• n = 150
• p = 155 or 0.15
• q = 1 - p

           = 1 - 0.15

           = 0.85

The Mean will be:

→ $\mu = np$

By substituting the values, we get

      $= 150\times 0.15$

      $= 22.5$

Standard deviation will be;

→ $\sigma = \sqrt{npq}$

      $= \sqrt{150\times 0.15\times 0.85}$

      $= 4.373$

By using the normal distribution to approximate the binomial distribution, we get

→ $P (X < 20) = P(X < 19.5)$

                    $= P(\frac{X-\mu}{\sigma} < \frac{19.5-22.5}{4.373} )$

                    $= P (Z < -0.69)$

                    $= 0.2451$

Thus the above answer i.e., option C) is correct.

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