Question

Problem F: The waiting time for clients at a dental clinic is normally distributed with an average waiting time of 16.2 minutes and a standard deviation of 3.4 minutes. (28) Refer to Problem F. What is the probability that a client spends less than 14 minutes waiting? (1) 0.1405 (2) 0.5011 (3) 0.2578 (4) 0.1240 (29) Refer to Problem F. What is the probability that a client spends more than 17.5 minutes waiting? (1) 0.6480 (2) 0.3520 (3) 0.5426 (4) 0.2363 (30) Refer to Problem F. What is the probability that a client spends between 15 and 17 minutes waiting? (1) 0.2316 (3) 0.5948 (2) 0.3632 (4) 0.1428

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)

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

   where z is standard normal variable

14 – 16.2 = P(Z < 3.4

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

= 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