Question

QUESTION 4 Consider that 1000 individuals will attend the theater described in Question #3. The show in the theater begins at 9:00 pm. Predict the number of people that are going to be late for the show.

question 3 attached

Answer 1

Q3) 8:30 pm is 30 minutes after 8pm while 8:55 pm is 55 minutes after 8 pm. We are given the distribution of X here:

$X \sim N(\mu, \sigma)$

From standard normal tables, we have:
P(Z < -0.126) = 0.45,
P(Z < 1.282) = 0.9

Using the above z scores to obtain the corresponding X values, we have here:

Mean - 0.126*SD = 30
Mean + 1.282*SD = 55

(1.282 - 0.126)SD = 55 - 30
1.156*SD = 25
SD = 21.63 minutes.

Therefore, Mean = 30 + 0.126SD = 30 + 0.126*21.63 = 32.7249

Therefore 32.7249 is the required mean value here.

Q4) We calculate the probability of time being more 60 minutes here as :
P(X > 60)

Converting it to a standard normal variable, we have here:

$P(Z > \frac{60 - 32.7249}{21.63})$

From standard normal tables, we get here:

Therefore expected number of people that would be late is computed here as:
= 0.1038*1000 = 0.1038*1000 = 103.8

Therefore 104 is the expected number here.