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The times taken by students to get up in the morning can be modelled by a normal distribution with mean 26.4 minutes and standard deviation 3.7 minutes.

a) For a random sample of 350 students, find the number who would be expected to take longer than 20 minutes to get up in the morning.

b) ‘Very slow’ students are students whose time to get up is more than 1.645 standard deviations above the mean. Find the probability that fewer than 3 students from a random sample of 8 students are ‘very slow’.

پاسخ تشریحی :
نمایش پاسخ

a) $P\left( {X \gt 20} \right) = P\left( {z \gt  - 6.4/3.7} \right)$

$ = P\left( {z \gt  - 1.730} \right)$

$ = 0.9582$

Number of students $ = 335$ or $336$

b) P(very slow) $ = 0.05$

$P\left( {0,{\text{ }}1,{\text{ }}2} \right) = $

${\left( {0.95} \right)^8} + {}^8{C_1}{\left( {0.05} \right)^1}{\left( {0.95} \right)^7} + {}^8{C_2}{\left( {0.05} \right)^2}{\left( {0.95} \right)^6}$

$ = 0.6634 + 0.2793 + 0.0515$

$ = 0.994$

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