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Solutions to Exercises 10
- Let X be the weight in kg of a bag of feed made in the mill. Then X ∼ N (8. 1 , 0. 072 ), that is, X has a normal distribution with mean μ = 8. 1 and standard deviation σ = 0. 07.
(a) The probability that the weight of a bag is over 8. 25 kg is P (X > 8 .25). Now P (X > 8 .25) = 1 − P (X ≤ 8 .25). To use the formula
P (X ≤ x) = P
Z ≤
x − μ σ
we need to calculate
z =
- 25 − μ σ
and from tables we obtain P (Z < 2 .14) = 0. 9838. A more accurate answer can be found using linear interpolation or by using Minitab. The accurate answer (using Minitab) is P (Z < 2 .14286) = 0. 9839 (to 4 decimal places). Therefore, using tables,
P (X > 8 .25) = 1 − 0 .9838 = 0. 0162.
The accurate answer is 0.0161. (b) The probability that the weight of a bag is between 8. 0 kg and 8. 25 kg is
P (8. 0 < X < 8 .25) = P (X < 8 .25) − P (X ≤ 8 .0).
From (a), using tables we have P (X < 8 .25) = 0. 9838. Also
z =
- 0 − μ σ
and from tables we obtain P (Z < − 1 .4286) = 0. 0766. Therefore
P (8. 0 < X < 8 .25) = 0. 9838 − 0. 0766 = 0. 9072.
Using Minitab, the accurate answer is P (8. 0 < X < 8 .25) = 0. 983938 − 0 .076563 =
- 907375 or 0.9074 to 4dp. (c) The probability that the weight of a bag is less than 8. 0 kg is P (X < 8 .0). Now
z =
- 0 − μ σ
and from tables we obtain P (Z < − 1 .4286) = 0. 0766. Therefore
P (X < 8 .0) = 0.0766 = 7.66%.
Therefore P (X ≥ 8 .0) = 1 − 0 .0766 = 0.9234 = 92.34%. So 92.34% can be used.
(d) We need the weight x so that P (X > x) = 0. 98. First, we need the value of z so that P (Z > z) = 0. 98 , that is P (Z < z) = 0. 02. Using tables, the two key probabilities are P (Z < − 2 .05) = 0. 0202 and P (Z < − 2 .06) = 0. 0197 and so we take
z = − 2 .06 +
× { 2. 06 − 2. 05 }
× 0. 01
In other words P (Z < − 2 .054) = 0. 02. Moving back to the weight scale, we need the value x such that P (X < x) = 0. 02 and so we take
x = μ + zσ = 8. 1 − 2. 054 × 0. 07 = 7. 956.
So 98% of the bags weigh more than 7.956kg.
- The amount of liquid discharged X has a normal distribution with mean μ = 200ml and standard deviation σ = 15ml.
(a) The probability that a cup contains less than 170 ml is P (X < 170). Now
z = 170 − μ σ
and from tables we obtain P (Z < −2) = 0. 0228. Therefore
P (X < 170) = 0.0228 = 2.28%.
(b) The probability that a cup contains over 225 ml is P (X > 225). Now P (X > 225) = 1 − P (X ≤ 225). Also
z = 225 − μ σ
and from tables we obtain P (Z < 1 .67) = 0. 9525. We also obtain P (Z < 1 .66) =
- Since 5 /3 = 1. 6667 is 2 / 3 of the way from 1.66 to 1.67 we use a probability which is 2 / 3 of the way from 0.9515 to 0.9525. That is, we use
0 .9515 +
× (0. 9525 − 0 .9515) = 0 .9515 +
× 0. 0010
Therefore P (X > 225) = 1 − 0 .9522 = 0.0478 = 4.78%. We can get the same answer using Minitab.
and from tables we obtain P (Z < − 2 .4) = 0. 0082 , and so P (X ≤ 10) = 0. 0082. Therefore
P (10 < X < 15) = 0. 3446 − 0. 0082 = 0. 3364.
(c) If only 3% of deliveries are late then we need the number of days x so that P (X ≤ x) = 0. 97. First, we need the value of z so that P (Z < z) = 0. 97. Using tables, the two key probabilities are
P (Z < 1 .88) = 0. 9699 and P (Z < 1 .89) = 0. 9706
and so we take
z = 1.88 +
× { 1. 89 − 1. 88 }
× 0. 01
In other words P (Z < 1 .8814) = 0. 97. Moving back to the delivery time scale, we need the value x such that P (X < x) = 0. 97 and so we take
x = μ + zσ = 16 + 1. 8814 × 2. 5 = 20. 70 days.
This is the same as the accurate answer (calculated using Minitab). (d) In the new processing system, X has a normal distribution with mean μ = 16 days and standard deviation σ = 1. 5 days. The probability of a delivery on time is P (X < 20). Now z = 20 − μ σ
and from tables we obtain P (Z < 2 .67) = 0. 9962. Therefore
P (X < 20) = 0. 9962.
This is an increase on the previous system of over 5%. This is the same as the exact answer calculated using Minitab.
- (a) Let the weight of 3 bananas be X 3. Then X 3 ∼ N (0. 45 , 0 .0075). The standard devia- tion is
0 .0075 = 0. 08660. Now
P (X 3 > 0 .5) = 1 − P (X 3 < 0 .5)
= 1 − P
Z <
= 1 − P (Z < 0 .57735)
From tables we find that P (Z < 0 .57) = 0. 7157 and P (Z < 0 .58) = 0. 7190. There- fore we calculate P (Z < 0 .57735) = 0.7157 + 0. 735 × (0. 7190 − 0 .7157) = 0. 7181. So P (X 3 > 0 .5) = 1 − 0 .7181 = 0. 2819. Minitab gives the same answer to 4dp. (b) Let the weight of 4 bananas be X 4. Then X 4 ∼ N (0. 60 , 0 .0100). The standard devia- tion is
0 .0100 = 0. 1. Now P (X 3 > 0 .5) = 1 − P (X 4 < 0 .5)
= 1 − P
Z <
= 1 − P (Z < −1)
From tables we find that P (Z < −1) = 0. 1379. So P (X 4 > 0 .5) = 1 − 0 .1379 =
Let the number of bananas which are needed to reach the weight of 0.5kg be N. We can see that P (X 3 > 0 .5) = P (N = 1) + P (N = 2) + P (N = 3) and P (X 4 > 0 .5) = P (N = 1) + P (N = 2) + P (N = 3) + P (N = 4). So P (N = 4) = P (X 4 > 0 .5) + P (X 3 > 0 .5) = 0. 8621 − 0 .2819 = 0. 5902. There is no lower limit to the value of a variable which has a normal distribution. Therefore, if the weights of bananas really had exactly a normal distribution, there would be a nonzero probability of finding a banana with a negative weight. This is, of course, impossible. How- ever, since zero is three standard deviations below the mean, the probability would be small and the normal distribution might be a reasonable approximation.
- Let the weight of a bag in g be X. The X ∼ N (1064, 502 ).
(a) A bag is underweight if X < 1000.
P (X < 1000) = P
Z <
= P (Z < − 1 .28)
(b) Let the number of underweight bags in a batch of 100 be N. Then N ∼ Bin(100, 0 .1003). Now np = 10. 03 and n(1 − p) = 89. 97 so we can use the normal approximation with μ = np = 10. 03 and σ^2 = np(1 − p) = 9. 023991. This gives a standard deviation σ =
So
P (N > 15) ≈ P
Z >
≈ P (Z > 1 .821)
≈ 1 − P (Z < 1 .821)
