Module 4 Problem Set 1-4 Answers
Problem Set 4.1 Solutions.
1. Solution.
The expected value is given by
So, the convenience store can expect to sell 8.48 boxes of sugar per week.
The variance is given by
The standa
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Module 4 Problem Set 1-4 Answers
Problem Set 4.1 Solutions.
1. Solution.
The expected value is given by
So, the convenience store can expect to sell 8.48 boxes of sugar per week.
The variance is given by
The standard deviation is given by:
2. Solution.
The expected value is given by
So, the barber can expect 2.46 customers per hour.
The variance is given by
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The standard deviation is given by:
3. Solution.
The expected value is given by
So, we can expect 1.7 broken bulbs per box.
Problem Set 4.2 Solutions
1 . Solution.
Here we have n=9, x=7, and p=.75.
2. Solution.
Here we have n=13, x=4, and p=.211.
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3. Solution.
For two or less, we must calculate the probability of getting two defective, one defective, and zero
defective, then add the probabilities.
For two defective, we have n=6, x=2, and p=.17.
For one defective, we have n=6, x=1, and p=.17.
For zero defective, we have n=6, x=0, and p=.17.
So, the probability of two, one, or zero defective:
.2057+.40178+.3269=.93438.
4. Solution.
The factory will be in danger if more than four safety systems fail. This means that the factory will be
in danger if five, six, or seven safety systems fail.
For five failures:
Here we have n=7, x=5, and p=.06.
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For six failures:
Here we have n=7, x=6, and p=.06.
For seven failures:
Here we have n=7, x=7, and p=.06.
The probability is given by:
.00001443+.000000307+.0000000028=.00001474.
Problem Set 4.3 Solutions
1. Find P(Z ≤ .35). This is asking for the probability that the standard normal random variable, Z, is
less than .35.
Recall that the standard normal distribution table gives us values to the left of the z value, so that is
exactly what we want. From the table, we get P(Z ≤ .35) = .63683.
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2. Find P(Z ≤ 1.04) . This is asking for the probability that the standard normal random variable, Z, is
less than 1.04.
The standard normal distribution table gives us values to the left of the z value, so that is exactly
what we want. From the table, we get P(Z ≤ 1.04) = .85083.
3. Find P(Z ≤ - .85). This is similar to the previous two problems. From the table, we get P(Z ≤ - .85)
= .19766.
4. Find P(Z ≤ 1.93). This is similar to the previous three problems. From the table, we get P(Z ≤
1.93) = .97320.
5. Find P(Z ≤ - .05). This is similar to the previous four problems. From the table, we get P(Z ≤ - .05)
= .48006.
6. Find P(Z ≥ .63) . This is asking for the area to the right of z=.63 (see the shaded area in the
diagram below) but if we look up z=.63 in the table, we will get the area to the left of z=.63. Since the
total area under the curve is 1, we can use:
P(Z ≥ .63) = 1 - P(Z ≤ .63)
Of course, P(Z ≤ .63) is given directly in the standard normal table and is found to be .73565. So,
P(Z ≥ .63) = 1 - P(Z ≤ .63) = 1 - .73565 = .26435.
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7. Find P(Z ≥ - 1.07). P(Z ≤ - 1.07) is given directly in the standard normal table and is found to be .
14231. So,
P(Z ≥ - 1.07) = 1 - P(Z ≤ - 1.07) =1 - .14231 = .85769
8. Find P(Z ≥ .36). P(Z ≤ .36). is given directly in the standard normal table and is found to be .
64058. So,
P(Z ≥ .36) = 1 - P(Z ≤ .36) = 1 - .64058 = .35942
9. Find P(Z ≥ 2.06). P(Z ≤ 2.06). is given directly in the standard normal table and is found to be .
98030. So,
P(Z ≥2.06) = 1 - P (Z ≤ 2.06) = 1 -.98030 =.0197
10. Find P(Z ≥ -.78). P(Z≤ -.78). is given directly in the standard normal table and is found to be .
21770. So,
P(Z ≥ - .78) = 1 - P(Z ≤ - .78) = 1 - .21770
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