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determine the probability?
The hours of sleep data for some students, suggest that the population of hours of sleep can be modeled as a normal distribution with mean = 7.2 hours and standard deviation = 1.3 hours.
(a) Determine the probability assigned to sleeping less than 6.5 hours.
(b) Find the 70th percentile of the distribution for hours of sleep.
2 Answers
- AlanLv 74 years ago
(a)
P(x<6.5) = P( Z< (X-mean)/standard deviation ) = P(z< (6.5-7.2) / 1.3 hours )
P(x< 6.5) =P ( Z< 0.7/1.3) = P (z < -0.538461538)
since -0.53846 is not exactly in a z-table, but
P(z< -0.54) = 0.29460
P(z< -0.53) = 0.29806
If we interpolate,
P(z< -0.53846) =+0.2946 + (-0.53846- (-0.54))* (0.29806-0.2946))/ 0.01
=0.29513284
(Round to 5 digits like the table)
P(z< -0.53846) = + 0.29513
P(x< 6.5) = 0.29513
(b) working on it
P(z< Z) = 0.70
Look up 0.70 inside of z-table
Two closest value are below.
P(z< 0.52) = .69847
P(z< 0.53)= .70194
Since it not really close to either value, I will interpolate
+0.52 + (0.70-0.69847)* (0.53-0.52)/ (0.70194-0.69847)
=0.524409222
Z = (x-mean)/ standard deviation
Z*standard deviation = x- mean
x = mean+ Z*standard deviaton
x = +7.2 + 0.524409222*1.3 = 7.881731989
x = 7.88173 hours