MGMT2262 Tutorial Sheet 4 Solutions

 

1)     P(X < 21) = (21 – 20.2)/(21.6 – 20.2) = 0.8/1.4 = 0.5714 = 57.14%

2)     P(X > 20.5) = (21.6 – 20.5)/(21.6 – 20.2) = 1.1/1.4 = 0.7857 = 78.57%

3)     m = (20.2 + 21.6)/2 = 20.9; s = (21.6 – 20.2)/sqrt(12) = 0.4041

m - 2s = 20.9 – 2(0.4041) = 20.9 – 0.8082 = 20.0918

m + 2s = 20.9 + 0.8082 = 21.7082

Notice that these limits lie outside the limits of 20.2 and 21.6. Since P(20.2 < X < 21.6) = 100%, it logically follows that P(20.0918 < X < 21.7082) = 100% as well. In fact, for any uniform distribution, 100% of the distribution always lies within 2 standard deviations of the mean, exceeding the 75% threshold under Chebyshev’s theorem.

Since the CSR serve 40 customers per hour, they serve 1 customer per 1.5 minutes on average.

4)     P(x > 2) = e-2/1.5 = e-1.333333 = 0.2636 = 26.36%

5)     P(x > 2 | x > 0.5) = P(x > 1.5) = e-1.5/1.5 = e-1 = 0.3679 = 36.79%

6)     If there are only 4 CSR instead of 5, they collectively serve 32 customers per hour on average or 1 customer per 1.875 minutes. Then P(x > 2) = e-2/1.875 = e-1.0666666 = 0.3442 = 34.42%

7)     P(x < 2 | x > 0.5) = 1 – P(x > 2 | x > 0.5) = 1 – P(x > 1.5) = 1 – e-1.5/1.875 =  1– e-0.8 = 1 – 0.4493 = 0.5507 = 55.07%

8)     Z = (35 – 30)/12.5 = 0.4. P(X < 35) = P(Z < 0.4) = 0.5 + 0.1554 = 0.6554.

9)     Z = (50 – 30)/12.5 = 1.6. P(X > 50) = P(Z > 1.6) = 0.5 - 0.4452 = 0.0548.

10)  P(25 < X < 50) = P(-0.4 < Z < 1.6).

P(-0.4 < Z < 0) = 0.1554 and P(0 < Z < 1.6) = 0.4452. Adding the two probabilities together, we get the same answer of 0.6006.

11)  Since we’re interesting in the top 10% of sales, we need the 90th percentile. Using the t table (across from the Z table inside the front cover), we find Z = 1.282. So X = 30 + 1.282(12.5) = 46.025 which rounds to 46.03.

12)  Since 2% is not a common tail probability, we need to work backward through the Z table. Finding 0.48 in the probability part of the Z table, we see that P(0 < Z < 2.05) = 0.4798 and P(0 < Z < 2.06) = 0.4803. Since 0.48 is approximately between 0.4798 and 0.4803, we take the Z value to be 2.055. Since we need the 2nd percentile, the needed Z value is –2.055. Then X = 30 –2.055(12.5) = 4.3125 which rounds to 4.31.