Answer:
Explanation:
(a) Calculate the approximate probability that between 35 and 70 tickets are given out on a particular day.
[tex]P(35\leq X\leq 70) =P(34.5\leq X \leq 70.5)[/tex] (using continuity correction)
[tex]=P(\frac{34.5- \mu}{\sqrt{\mu} } \leq \frac{X - \mu}{\sqrt{\mu} } \leq \frac{70.5-\mu}{\sqrt{\mu} } )\\\\=P(\frac{34.5-50}{\sqrt{50} } \leq Z \leq\frac{70.5-50}{\sqrt{50} } \\\\=P(-2.19\leq Z \leq 2.90)\\\\=P(Z\leq 2.90)-P(Z\leq -2.19)[/tex]
= 0.9981 - 0.0143 (using standard normal tables)
= 0.9839
b) Consider [tex]X_t[/tex] is the random variable that represents the number of parking tickets issued in certain city in a 5-day week
The mean number of parking ticket issued in a particular city on 5-day week is
[tex]\mu_t = \mu *t\\\\=5 \times 50 = 250[/tex]
Therefore, the required probability is
[tex]P(225\leq X_t \leq 275)=(224.5\leq X_t\leq 275.5)\\\\=P(\frac{224.5-\mu_t}{\sqrt{\mu_t} } \leq \frac{X_t-\mu_t}{\sqrt{\mu_t} } \leq \frac{275.5-\mu_t}{\sqrt{\mu_t} } )\\\\=P(\frac{224.5-250}{\sqrt{250} } \leq Z \leq \frac{275.5-250}{\sqrt{250} } \\\\=P(-1.61\leq Z\leq 1.61)\\\\=P(Z\leq 1.61)-P(Z\leq -1.61)[/tex]
= 0.9463 - 0.0537 (using standard normal tables)
= 0.8926
c) see the attached file
image 1 (the approximate probability of part a )
image 2 (the approximate probability of part b )
From the two graph observed that the probability obtained using software is approximately same as calculated by manually
