Question

t (even as simple as 11 - 5 6, if you think it, write it randomly ou may tor shortcut here, but if you do any work beyond t immed may use your calculator shortcut here, but if you selected cars. Use this edate display, show i The table contains the results of the spins for ten sample data to find the following Remember, if grey slot they tay land on a white slot they get to ride the golf cart and if they land on a they have to walk. Car # | Status | Level Parts A C apply to the garage level 1 Walk 2 Walk 3 Walk 4 Walk 5 Ride 6 Walk 7 Walk 8 Ride 9 Walk 10 Walk Mean A) 6 15 9 15 9 9 B) Standard Deviation c) Variance 15 15 6 9 D) Frequency of Rides 11 Ride proportion of Rides 12 Walk 3) When drivers spin The Wheel of Parking" it spins for 42 seconds on average with a standard deviation of 4 seconds. Legend has it that when Danny Ocean spun the wheel it took 75 seconds to come to a stop which is more than 8 standard deviations longer than the average spin. A) What does Chebyshev's inequality give as the maximum probability of the time the wheel spins being more than 8 standard deviations away from the mean of 42 seconds? B) Assuming that the time the wheel spins is symmetrically distributed approximately 95% of all spins would last between and _seconds

Answer 1

2)

A)

Mean of Garage levels = (2 + 6 + 15 + 9 + 15 + 9 + 9 + 6 + 15 + 15 + 6 + 9) / 12 = 116/12 = 9.666667

B)

Standard deviation = sqrt[[(2 - 9.666667)² + (6 - 9.666667)² + (15 - 9.666667)² + (9 - 9.666667)² + (15 - 9.666667)² + ( 9 - 9.666667)² + (9 - 9.666667)² + (6 - 9.666667)² + (15 - 9.666667)² + (15 - 9.666667)² + (6 - 9.666667)² + (9 - 9.666667)² ] / (12-1)]

= sqrt[(58.7777829 + 13.4444469 + 28.4444409 + 0.4444449 + 28.4444409 + 0.4444449 + 0.4444449 + 13.4444469 + 28.4444409 + 28.4444409 + 13.4444469 + 0.4444449) / 11]

= sqrt(214.6667 /11)

= sqrt(19.51515)

= 4.417595

c)

Variance = Standard deviation² = 4.417595² = 19.51515

D)

From the data, Frequency of Rides = 3

E)

Proportion of Rides = 3/12 = 0.25

3.

A.

By Chebyshev's inequality,

$P\left ( |X - EX| \ge b \right ) \le Var(X)/b^2$

Given, the time wheel spins more than 8 standard deviations away from the mean.

Then putting $b = 8 \sigma$ in the above equation, we get

$P\left ( |X - EX| \ge 8 \sigma \right ) \le \sigma^2/(8 \sigma)^2$

$P\left ( |X - EX| \ge 8 \sigma \right ) \le 1/64$

Thus, the maximum probability that the time wheel spins more than 8 standard deviations away from the mean is 1/64 = 0.015625

B)

z value for 95% probability is $\pm 1.96$

So, the lower limit = 42 - 1.96 * 4 = 34.16

upper limit = 42 + 1.96 * 4 = 49.84

Thus, approximately 95% of all spins would last between 34.16 and 49.84 seconds.