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Problem F: The waiting time for clients at a dental clinic is normally distributed with an average waiting time of 16.2 minut

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Answer #1

Define random variable X: Waiting time for clients at a dental clinic

X is normally distributed with mean = \mu = 16.2 and standard deviation = \sigma = 3.4

28)

Here we have to find P(X < 14)

P(X<14)=P(\frac{X-\mu}{\sigma}<\frac{14-\mu}{\sigma})

  =P(z<\frac{14-16.2}{3.4}) where z is standard normal variable

=P(z<\frac{-2.2}{3.4})

= P(z < -0.65) (Round to 2 decimal)

= 0.2578 (From statistical table of z values)

The probability that a client spends less than 14 minutes waiting is 0.2578

29)

Here we have to find P(X > 17.5)

P(X>17.5)=P(\frac{X-\mu}{\sigma}>\frac{17.5-\mu}{\sigma})

  =P(z>\frac{17.5-16.2}{3.4}) where z is standard normal variable

=P(z>\frac{1.3}{3.4})

= P(z > 0.38) (Round to 2 decimal)

= 1 - P(z < 0.38)

= 1 - 0.6480 (From statistical table of z values)

= 0.3520

The probability that a client spends more than 17.5 minutes waiting is 0.3520

30)

Here we have to find P(15 < X < 17)

P(15<X<17)=P(\frac{15-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{17-\mu}{\sigma})

  =P(\frac{15-16.2}{3.4}<z<\frac{17-16.2}{3.4}) where z is standard normal variable

  =P(\frac{-1.2}{3.4}<z<\frac{0.8}{3.4})

= P(-0.35 < z < 0.24) (Round to 2 decimal)

= P(z < 0.24) - P(z < -0.35)

= 0.5948 - 0.3632 (From statistical table of z values)

= 0.2316

The probability that a client spends between 15 and 17 minutes waiting is 0.2316

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