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میتونی لایو بذاری!

The times spent by people visiting a certain dentist are independent and normally distributed with a mean of 8.2 minutes. 79% of people who visit this dentist have visits lasting less than 10 minutes.

a) Find the standard deviation of the times spent by people visiting this dentist.

b) Find the probability that the time spent visiting this dentist by a randomly chosen person deviates from the mean by more than 1 minute.

c) Find the probability that, of 6 randomly chosen people, more than 2 have visits lasting longer than 10 minutes.

d) Find the probability that, of 35 randomly chosen people, fewer than 16 have visits lasting less than 8.2 minutes.

پاسخ تشریحی :
نمایش پاسخ

a) $z = 0.807$

$0.807 = \frac{{10 - 8.2}}{\sigma }$

$s = 2.23$

b) $P$($ \gt 1$ min from mean) $ = P$(mod $z \gt \frac{1}{{2.23}}$)

$ = P\left( {\left| z \right| \gt 0.4484} \right)$

$ = \left( {1 - 0.6729} \right) \times 2$

$ = 0.654$

c) $P$($ \gt 2$ longer) $ = 1 - P$($0$, $1$, $2$ longer)

$ = 1 - \{ {\left( {0.79} \right)^6} + {}^6{C_1}\left( {0.21} \right)\left( {0.79} \right)5 + $

${}^6{C_2}{\left( {0.21} \right)^2}{\left( {0.79} \right)^4}\} $

$ = 0.112$

d) $\mu  = 35 \times 0.5 = 17.5$

${\sigma ^2} = 35 \times 0.5 \times 0.5 = 8.75$

$P\left( {X \lt 16} \right) = \Phi \left( {\frac{{15.5 - 17.5}}{{\sqrt {8.75} }}} \right)$

$ = 1 - \Phi \left( {0.676} \right)$

$ = 1 - 0.7505$

$ = 0.2495{\text{ }}\left( {0.249{\text{ }}or{\text{ }}0.250} \right)$

OR ${}^{35}{C_0}{0.5^0}{0.5^{35}} + {}^{35}{C_1}{0.5^1}{0.5^{34}} + {}^{35}{C_2}{0.5^2}{0.5^{33}} + ...$

$ = 8582372584/{2^{35}} = 0.250$

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