Download z-score practice answers

Normal Distribution and Z scores
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1.
You purchased 10 baskets of strawberries at the local farmer's market and counted the number of
strawberries in each basket. Based on your purchases, do you think the number of strawberries in a
basket is normally distributed?
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(A)Yes
(B)No
(C)Not enough information
(D)All of the above
Explanation:
All of the above? How is that even possible? This data is normally distributed. The first way to
determine this is to calculate the mean and the median, which are both equal to 20. This is a big clue that
the data is normally distributed. But, just to make sure, we should check to make sure we are dealing
with a normal (Liberty Bell) curve. In this case, we can see that the mound will occur at 20.
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2.
Andy Lee is the punter for the San Francisco 49ers. He had a stellar 2011 season with an average punt
length of 50.9 yards with a standard deviation of 3.5. His punt distance follows a normal distribution.
Determine the range of punt distances that covers 68% of the distances.
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(A)49.9 to 50.9 yards
(B)48.9 to 51.9 yards
(C)47.4 to 54.4 yards
(D)43.9 to 57.9 yards
Explanation:
Because this data follows a normal distribution, we can use the empirical rule. We know that 68% of the
data will be within one standard deviation of the mean in both the positive and negative directions. The
mean is an amazing 50.9 yards with a decent standard deviation of 3.5 yards. So one standard deviation
below is 50.9 – 3.5 = 47.4 and one standard deviation above is 50.9 + 3.5 = 54.4.
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3.
Andy Lee is the punter for the San Francisco 49ers. He had a stellar 2011 season with an average punt
length of 50.9 yards with a standard deviation of 3.5. His punt distance follows a normal distribution.
Determine interval that contains 95% of data.
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(A)49.9 to 50.9 yards
(B)48.9 to 51.9 yards
(C)47.4 to 54.4 yards
(D)43.9 to 57.9 yards
Explanation:
Because this data follows a normal distribution, we can use the empirical rule. We know that 95% of the
data will be within two standard deviations of the mean in both the positive and negative directions. The
mean is an amazing 50.9 yards with a decent standard deviation of 3.5 yards. So two standard deviations
below is 50.9 – 2(3.5) = 43.9 and two standard deviations above is 50.9 + 2(3.5) = 57.9.
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4.
Andy Lee is the punter for the San Francisco 49ers. He had a stellar 2011 season with an average punt
length of 50.9 yards with a standard deviation of 3.5. His punt distance follows a normal distribution. In
the very last game of the post-season, Andy Lee made his last punt for 39 yards. What is the Z-score for
this punt?
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(A)-3.4
(B)3.4
(C)-11.9
(D)5.5
Explanation:
Remember that the Z-score, which also normalizes the value is Z = x- μ⁄σ. In this
case,
. The sign is important! The negative Z-score tells us the value is less than
the mean (rather than greater than the mean).
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5.
Andy Lee is the punter for the San Francisco 49ers. He had a stellar 2011 season with an average punt
length of 50.9 yards with a standard deviation of 3.5. His punt distance follows a normal distribution.
Andy Lee's first punt of the season was 66 yards. What is the Z-score for this punt?
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(A)-2.4
(B)2.4
(C)4.3
(D)5.5
Explanation:
Remember that the Z-score, which also normalizes the value is Z = x- μ⁄σ. In this case, Z = 66 - 50.9⁄3.5 = 4.3.
The sign is important! The positive Z-score tells us the value is greater than the mean, which we know is
true because 66 > 50.9.

6.
The length of a phone conversation, measured in minutes, follows a normal distribution with a mean of
7 minutes and a standard deviation of 2.2 minutes. What is the probability that the phone conversation
lasts less than 8.5 minutes?
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(A)0.75
(B)0.80
(C)0.25
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(D)1.0
Explanation:
Logically, we know that some conversations last more than 8.5 minutes, so (D) can't be right. We also
know that the majority of phone conversations last 8.5 minutes or below, so (C) can't be right, either. To
determine the probability that a conversation is less than or equal to 8.5 minutes, first we need to
determine the Z-score of 8.5 minutes which is Z = 8.5-7.0⁄2.2 = 0.68. Using a standard normal table reveals
that this is associated with a probability of 0.75, so the correct answer is (A).
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7.
The length of a phone conversation, measured in minutes, follows a normal distribution with a mean of
7 minutes and a standard deviation of 2.2 minutes. What is the probability that the phone conversation
lasts more than 15 minutes?
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(A)0.75
(B)0.00014
(C)0.014
(D)1.0
Explanation:
To determine the probability that a conversation is greater than 15 minutes, we need to determine the
probability that the phone conversation is less than or equal to 15 and then subtract that value from 1.0.
The Z-score table reveals that the probability that a conversation is less than or equal to 15 minutes is
0.9996, so the probability that the conversation is greater than 15 minutes is 0.00014, or (B). This makes
sense, since the probability of a long phone conversation is very small, meaning definitely not (A) or
(D).
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8.
The length of a phone conversation, measured in minutes, follows a normal distribution with a mean of
7 minutes and a standard deviation of 2.2 minutes. What is the probability that the phone conversation
lasts between 8.5 and 15 minutes?
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(A)0.75
(B)0.80
(C)0.2498
(D)1.0
Explanation:
To determine the probability that a conversation lies between the two values, we determine the
probability that the conversation is less than or equal to 15 minutes and subtract the probability that it is
less than or equal to 8.5 minutes. This is 0.9996 – 0.75 = 0.2498. We could also consider this logically,
knowing that (A) can't be right because the number of conversations less than 8.5 minutes is already
0.75 (meaning the probability of conversations over 8.5 minutes can't exceed 0.25), (B) can't be right for
the same reason, and (D) makes no sense since some conversations are under 8.5 minutes, and some are
over 15.

9.
The class mean for a recent chemistry exam was 80.5 with a standard deviation of 4.2. What is the Zscore of a student who receives an 87 on the exam?
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(A)1.54
(B)-1.54
(C)2.0
(D)5.0
Explanation:
We can calculate the Z-score using the formula Z = x- μ⁄σ = 87-80.5⁄4.2 = 1.54. In other words, the student's
score of 87 is 1.54 standard deviations away from the average score in the class. Since the Z-score is
positive, that means the value is greater than the average.

10.
The class mean for a recent chemistry exam was 85.5 with a standard deviation of 2.2. What is the Zscore of a student who receives an 87 on this exam?
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(A)-0.68
(B)-0.2
(C)0.2
(D)0.68