Data Analysis Alıştırma Soruları — GRE General Test — Sayfa 9

Soru 161Soru →

In a sample space of a random experiment, $A$ and $B$ are independent events such that $P(A) = 0.35$ and $P(A \cup B) = 0.74$ . Event $C$ is mutually exclusive with Event $A$ . If the conditional probability $P(C \mid B) = 0.20$ , what is the probability that Event $B$ occurs, but neither Event $A$ nor Event $C$ occurs?

0.19

0.27

0.39

0.48

0.51

Cevabı ve açıklamayı göster

Cevap: 0.27

Cevap

0.27

To find the probability that Event

B

occurs without

A

or

C

, we must isolate the region of

B

that does not overlap with

A

or

C

. Since

A

and

B

are independent,

P(A \cup B) = P(A) + P(B) - P(A)P(B)

, which allows us to solve for

P(B) = 0.60

and

P(A \cap B) = 0.21

. Next, using the conditional probability

P(C \mid B) = 0.20

, we find

P(B \cap C) = 0.20 \times 0.60 = 0.12

. Because

A

and

C

are mutually exclusive, the intersections

A \cap B

and

B \cap C

do not overlap. Subtracting both intersection probabilities from

P(B)

yields

0.60 - 0.21 - 0.12 = 0.27

.

Adım Adım Çözüm

1

Calculate the probability of Event

B

,

P(B)

, using the independence of

A

and

B

.

P(B) = 0.60

Since

A

and

B

are independent,

P(A \cap B) = P(A)P(B)

. Substituting into the union formula

P(A \cup B) = P(A) + P(B) - P(A)P(B)

gives

0.74 = 0.35 + P(B)(1 - 0.35)

, so

0.39 = 0.65 P(B)

, yielding

P(B) = 0.60

.

2

Find the joint probability

P(A \cap B)

.

P(A \cap B) = 0.21

By independence,

P(A \cap B) = P(A) \times P(B) = 0.35 \times 0.60 = 0.21

.

3

Calculate the joint probability

P(B \cap C)

using the conditional probability formula.

P(B \cap C) = 0.12

From the definition of conditional probability,

P(C \mid B) = \frac{P(B \cap C)}{P(B)}

, so

P(B \cap C) = P(C \mid B) \cdot P(B) = 0.20 \times 0.60 = 0.12

.

4

Determine the probability that

B

occurs but neither

A

nor

C

occurs.

P(B \cap A^c \cap C^c) = 0.27

Since

A

and

C

are mutually exclusive, events

(A \cap B)

and

(B \cap C)

are disjoint subsets of

B

. Therefore,

P(B \cap A^c \cap C^c) = P(B) - P(A \cap B) - P(B \cap C) = 0.60 - 0.21 - 0.12 = 0.27

.

Anahtar Kavram

Probability Rules for Independent, Dependent, and Mutually Exclusive Events

Soru 162Soru →

A jar contains $4$ red marbles, $6$ blue marbles, and $5$ green marbles. If two marbles are selected at random one after another without replacement, what is the probability that the first marble selected is red and the second marble selected is blue?

\frac{4}{35}

\frac{8}{75}

\frac{73}{105}

\frac{12}{55}

\frac{10}{29}

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{35}$

Cevap

The probability that the first marble selected is red and the second marble selected is blue is

\frac{4}{35}

.

To find the joint probability of two dependent sequential events, multiply the probability of the first event by the conditional probability of the second event. The probability of drawing a red marble first is

\frac{4}{15}

. Because the selection is made without replacement, there are

14

marbles remaining in the jar,

6

of which are blue. The probability of drawing a blue marble second is

\frac{6}{14}

. Multiplying these together yields

\frac{4}{15} \times \frac{6}{14} = \frac{24}{210} = \frac{4}{35}

.

Adım Adım Çözüm

1

Calculate the total number of marbles in the jar initially.

Total marbles =

4 + 6 + 5 = 15

.

The sample space size for the first draw is the sum of all marbles.

2

Determine the probability of selecting a red marble on the first draw.

P(\text{Red}_1) = \frac{4}{15}

.

There are

4

favorable outcomes out of

15

total outcomes.

3

Determine the probability of selecting a blue marble on the second draw given that one red marble was removed without replacement.

P(\text{Blue}_2 \mid \text{Red}_1) = \frac{6}{14} = \frac{3}{7}

.

After removing one red marble, there are still

6

blue marbles, but only

14

total marbles remaining.

4

Apply the multiplication rule for dependent events.

P(\text{Red}_1 \text{ and } \text{Blue}_2) = \frac{4}{15} \times \frac{6}{14} = \frac{24}{210} = \frac{4}{35}

.

The joint probability of sequential dependent events is the product of the first event's probability and the conditional probability of the second event.

Anahtar Kavram

Conditional Probability and Dependent Events

Tahmini Süre:1m 30s

Soru 163Soru →

A quality control manager records the durability ratings, on a scale from $1$ to $50$ , for $8$ randomly selected component batches. The ratings are listed below:

34, 18, 42, 27, 18, 39, 45, 25

Two additional component batches with identical ratings, $x$ and $y$ (where $x = y$ ), are included in the dataset. If the median rating of the updated dataset of $10$ batches is equal to the median rating of the original dataset of $8$ batches, what is the value of $x$ ?

22.5

26.0

30.5

31.5

36.5

Cevabı ve açıklamayı göster

Cevap: $30.5$

Cevap

The value of

x

is

30.5

.

To find the median of the original dataset, the numbers must first be ordered from smallest to largest:

18, 18, 25, 27, 34, 39, 42, 45

. Since there are

8

numbers, the median is the average of the 4th and 5th numbers:

(27 + 34)/2 = 30.5

. When two equal values

x

are added to form a

10

-element dataset, having

x = 30.5

places both new values directly between

27

and

34

. The 5th and 6th terms of the updated set are both

30.5

, making the new median

(30.5 + 30.5)/2 = 30.5

, which matches the original median.

Adım Adım Çözüm

1

Sort the original dataset of 8 durability ratings in ascending order.

The sorted dataset is

18, 18, 25, 27, 34, 39, 42, 45

.

Finding the median of a numerical dataset requires arranging the numbers in order from least to greatest.

2

Calculate the median of the original 8 ratings.

The 4th value is

27

and the 5th value is

34

. The median is

\frac{27 + 34}{2} = 30.5

.

For a dataset with an even number of elements (

n = 8

), the median is the average of the two middle elements (the 4th and 5th terms).

3

Analyze how adding two identical values

x

and

y

(

x = y

) affects the median of the 10-element dataset.

Inserting two values equal to

30.5

places them as the 5th and 6th elements in the 10-element sorted array:

18, 18, 25, 27, 30.5, 30.5, 34, 39, 42, 45

. The new median is

\frac{30.5 + 30.5}{2} = 30.5

.

To keep the median unchanged at

30.5

when adding two identical values, the added values must fall between

27

and

34

and specifically equal

30.5

so that the middle two terms of the 10 values average to

30.5

.

Anahtar Kavram

Median of a Dataset

Tahmini Süre:1m 30s

Soru 164Soru →

The daily water consumption per household in a municipal district is normally distributed with a mean of $320$ liters and a standard deviation of $40$ liters. A local utility company classifies households into three consumption categories:

- Efficient: Daily consumption below $240$ liters
- Moderate: Daily consumption between $240$ liters and $400$ liters
- High: Daily consumption above $400$ liters

Based on the $68\text{--}95\text{--}99.7$ empirical rule for normal distributions, in a random sample of $10,000$ households, how many more households are classified as Moderate than are classified as Efficient?

5,200

7,900

9,000

9,250

9,500

Cevabı ve açıklamayı göster

Cevap: $9,250$

Cevap

9,250

households

The boundary values

240

liters and

400

liters represent

z = -2

and

z = +2

, respectively. According to the empirical rule,

95\%

of the population falls between these limits, yielding

9,500

Moderate households. The remaining

5\%

is split equally into the upper and lower tails (

2.5\%

each), meaning

2.5\%

(

250

households) are Efficient. Subtracting

250

from

9,500

yields

9,250

.

Adım Adım Çözüm

1

Calculate the z-scores for the boundaries of the Moderate and Efficient categories.

The boundary

240

liters corresponds to

z = \frac{240 - 320}{40} = -2.0

. The boundary

400

liters corresponds to

z = \frac{400 - 320}{40} = +2.0

.

Standardizing values into z-scores allows the direct application of the empirical rule.

2

Determine the percentage and count of households in the Moderate category.

By the

68\text{--}95\text{--}99.7

rule,

95\%

of the data lies between

z = -2.0

and

z = +2.0

. Count =

0.95 \times 10,000 = 9,500

households.

The Moderate category spans from

\mu - 2\sigma

to

\mu + 2\sigma

.

3

Determine the percentage and count of households in the Efficient category.

The area below

z = -2.0

is

\frac{100\% - 95\%}{2} = 2.5\%

. Count =

0.025 \times 10,000 = 250

households.

The normal distribution is symmetric, so the remaining

5\%

outer area is evenly split into two tails of

2.5\%

each.

4

Calculate the difference between Moderate and Efficient household counts.

9,500 - 250 = 9,250

households.

Subtracting the count of Efficient households from Moderate households satisfies the core question prompt.

Anahtar Kavram

Normal distribution z-score standardization and asymmetric region calculations using the 68-95-99.7 empirical rule.

Tahmini Süre:2m 30s

Soru 165Soru →

A research laboratory conducted $6$ experimental trials to measure the duration, in milliseconds, of a specific chemical reaction. The durations recorded for $5$ of the trials were $240$ , $215$ , $260$ , $225$ , and $245$ . If the median duration of all $6$ trials was $235$ milliseconds, what was the duration, in milliseconds, of the 6th trial?

Cevabı ve açıklamayı göster

Cevap: 230

Cevap

230

For a dataset with 6 numbers, the median is the arithmetic mean of the 3rd and 4th numbers in ascending order. Arranging the 5 given numbers gives

215, 225, 240, 245, 260

. Since the target median is

235

, the sum of the two middle numbers must be

235 \times 2 = 470

. Placing

x = 230

into the dataset yields the ordered set

215, 225, 230, 240, 245, 260

, where the 3rd and 4th numbers are

230

and

240

. Their mean is

(230 + 240) / 2 = 235

, which matches the given median.

Adım Adım Çözüm

1

Order the 5 given numbers from least to greatest

The sorted list of known values is

215, 225, 240, 245, 260

.

Calculating or using median requires ordering the data points.

2

Express the median condition for an even number of data points (

n = 6

)

\text{Median} = \frac{\text{3rd value} + \text{4th value}}{2} = 235

, so

\text{3rd value} + \text{4th value} = 470

.

For an even number of values, the median is the average of the two middle numbers.

3

Determine the position and value of the unknown 6th trial

x

If

x \le 225

, the 3rd and 4th values would be

225

and

240

(median

232.5

). If

x \ge 245

, the 3rd and 4th values would be

240

and

245

(median

242.5

). Thus,

x

must lie between

225

and

240

.

Analyzing boundary conditions places

x

as the 3rd value and

240

as the 4th value.

4

Solve for

x

x + 240 = 470 \implies x = 230

.

The sum of the two middle values must equal

470

to yield a median of

235

.

Anahtar Kavram

Finding a missing value in a dataset given the median of an even number of observations.

Soru 166Soru →

A reliability engineering test evaluates two independent components, Component X and Component Y, in a machine. The probability that Component X fails during operation is $0.20$ , and the probability that Component Y fails during operation is $0.30$ . What is the probability that at least one of the two components operates successfully during operation?

0.44

0.50

0.56

0.70

0.94

Cevabı ve açıklamayı göster

Cevap: $0.94$

Cevap

The probability that at least one of the two components operates successfully is

0.94

.

The correct answer is

0.94

. The complement of the event 'at least one component operates successfully' is the event 'both components fail'. Because Component X and Component Y fail independently,

P(\text{both fail}) = P(\text{X fails}) \times P(\text{Y fails}) = 0.20 \times 0.30 = 0.06

. Subtracting this complementary probability from

1

yields

1 - 0.06 = 0.94

.

Adım Adım Çözüm

1

Determine the probability that each component fails.

P(\text{X fails}) = 0.20

and

P(\text{Y fails}) = 0.30

.

These probabilities are explicitly given in the problem statement.

2

Calculate the joint probability that BOTH components fail simultaneously using the multiplication rule for independent events.

P(\text{both fail}) = P(\text{X fails}) \times P(\text{Y fails}) = 0.20 \times 0.30 = 0.06

.

Since the components fail independently, their joint failure probability is the product of their individual failure probabilities.

3

Apply the complement rule to find the probability that at least one component operates successfully.

P(\text{at least one succeeds}) = 1 - P(\text{both fail}) = 1 - 0.06 = 0.94

.

The event 'at least one component succeeds' is the exact complement of 'both components fail'.

Anahtar Kavram

Independent Events and Complement Probability Rule

Soru 167Soru →

An executive is monitoring two independent corporate projects, Project Alpha and Project Beta. Based on historical performance, the probability that Project Alpha meets its deadline is $\frac{4}{5}$ , and the probability that Project Beta meets its deadline is $\frac{3}{4}$ . What is the probability that exactly one of the two projects meets its deadline?

\frac{7}{20}

\frac{3}{5}

\frac{19}{20}

\frac{7}{9}

\frac{1}{20}

Cevabı ve açıklamayı göster

Cevap: $\frac{7}{20}$

Cevap

The probability that exactly one project meets its deadline is

\frac{7}{20}

.

The event 'exactly one project meets its deadline' consists of two mutually exclusive scenarios: (1) Alpha meets its deadline and Beta does not, or (2) Alpha misses its deadline and Beta meets it. Using independence, the probability of Scenario 1 is

\frac{4}{5} \times \left(1 - \frac{3}{4}\right) = \frac{4}{5} \times \frac{1}{4} = \frac{4}{20}

. The probability of Scenario 2 is

\left(1 - \frac{4}{5}\right) \times \frac{3}{4} = \frac{1}{5} \times \frac{3}{4} = \frac{3}{20}

. Summing these mutually exclusive probabilities gives

\frac{4}{20} + \frac{3}{20} = \frac{7}{20}

.

Adım Adım Çözüm

1

Determine the probabilities of individual events and their complements.

P(\text{Alpha meets}) = \frac{4}{5}

,

P(\text{Alpha misses}) = 1 - \frac{4}{5} = \frac{1}{5}

.

P(\text{Beta meets}) = \frac{3}{4}

,

P(\text{Beta misses}) = 1 - \frac{3}{4} = \frac{1}{4}

.

To find the probability of specific outcome combinations, the complementary probabilities for each independent event are required.

2

Identify the mutually exclusive cases that satisfy the condition 'exactly one project meets its deadline'.

Case 1: Alpha meets and Beta misses. Case 2: Alpha misses and Beta meets.

The event 'exactly one' consists of two distinct, non-overlapping scenarios.

3

Calculate the joint probability for each case using independence.

Case 1 probability:

\frac{4}{5} \times \frac{1}{4} = \frac{4}{20}

. Case 2 probability:

\frac{1}{5} \times \frac{3}{4} = \frac{3}{20}

.

Since the projects operate independently, joint probabilities are found by multiplying individual event probabilities.

4

Add the probabilities of the mutually exclusive cases.

\frac{4}{20} + \frac{3}{20} = \frac{7}{20}

.

For mutually exclusive events, the total probability of either case occurring is the sum of their individual probabilities.

Anahtar Kavram

Probability of Independent Events and Mutually Exclusive Cases

Soru 168Soru →

A laboratory tests two solar panels, Panel A and Panel B, under identical conditions. The probability that Panel A operates at peak efficiency on any given day is $0.75$ , and the probability that Panel B operates at peak efficiency on any given day is $0.60$ . The daily efficiency outcomes of the two panels are independent events. Which of the following statements must be true? Select all such statements.

Geçerli olan tümünü seçin

The probability that both panels operate at peak efficiency on a given day is

0.45

.

The probability that neither panel operates at peak efficiency on a given day is

0.35

.

The probability that at least one panel operates at peak efficiency on a given day is

0.90

.

The conditional probability that Panel A operates at peak efficiency, given that Panel B operates at peak efficiency, is

0.45

.

The probability that Panel A operates at peak efficiency and Panel B does not operate at peak efficiency on a given day is

0.30

.

Cevabı ve açıklamayı göster

Cevap: The probability that both panels operate at peak efficiency on a given day is $0.45$ .; The probability that at least one panel operates at peak efficiency on a given day is $0.90$ .; The probability that Panel A operates at peak efficiency and Panel B does not operate at peak efficiency on a given day is $0.30$ .

Cevap

The correct statements are: the probability that both panels operate at peak efficiency is

0.45

; the probability that at least one panel operates at peak efficiency is

0.90

; and the probability that Panel A operates at peak efficiency while Panel B does not is

0.30

.

The statements confirming that both panels operate at peak efficiency (

0.45

), that at least one operates at peak efficiency (

0.90

), and that Panel A operates while Panel B does not (

0.30

) are mathematically sound applications of independent event rules.

Adım Adım Çözüm

1

Identify given probabilities and independence condition

P(A) = 0.75

,

P(B) = 0.60

, and events

A

and

B

are independent.

Establishes the given parameter values.

2

Calculate joint probability of both events occurring

P(A \cap B) = P(A) \times P(B) = 0.75 \times 0.60 = 0.45

.

For independent events, joint probability equals the product of individual probabilities.

3

Determine complement probabilities and probability of neither event occurring

P(A^c) = 1 - 0.75 = 0.25

,

P(B^c) = 1 - 0.60 = 0.40

, so

P(A^c \cap B^c) = 0.25 \times 0.40 = 0.10

.

Complements of independent events are also independent.

4

Calculate the union probability (at least one panel at peak efficiency)

P(A \cup B) = 1 - P(A^c \cap B^c) = 1 - 0.10 = 0.90

.

The event 'at least one' is the logical complement of 'neither'.

5

Evaluate conditional probability

P(A \mid B)

and difference probability

P(A \cap B^c)

P(A \mid B) = P(A) = 0.75

and

P(A \cap B^c) = 0.75 \times 0.40 = 0.30

.

Independence implies

P(A \mid B) = P(A)

and

P(A \cap B^c) = P(A) P(B^c)

.

Anahtar Kavram

Probability rules for independent events, complement rule, and conditional probability definition

Tahmini Süre:1m 30s

Soru 169Soru →

Events $A$ and $B$ are mutually exclusive, with $P(A) = 0.25$ and $P(B) = 0.40$ . Event $C$ is independent of both event $A$ and event $B$ , with $P(C) = 0.50$ . What is the probability that event $C$ occurs and at least one of events $A$ or $B$ occurs?

0.050

0.125

0.325

0.650

0.825

Cevabı ve açıklamayı göster

Cevap: $0.325$

Cevap

The probability that event

C

occurs and at least one of events

A

or

B

occurs is

0.325

.

The correct answer is

0.325

. First, since

A

and

B

are mutually exclusive events, the probability of at least one of them occurring is

P(A \text{ or } B) = P(A) + P(B) = 0.25 + 0.40 = 0.65

. Second, because event

C

is independent of both events, the probability that

C

occurs AND at least one of

A

or

B

occurs is given by the multiplication rule for independent events:

P(C) \times P(A \text{ or } B) = 0.50 \times 0.65 = 0.325

.

Adım Adım Çözüm

1

Calculate the probability of the union of mutually exclusive events

A

and

B

.

P(A \text{ or } B) = P(A) + P(B) = 0.25 + 0.40 = 0.65

Since

A

and

B

are mutually exclusive,

P(A \cap B) = 0

, so their combined probability is simply the sum of their individual probabilities.

2

Calculate the joint probability of event

C

and event

(A \text{ or } B)

.

P(C \text{ and } (A \text{ or } B)) = P(C) \times P(A \text{ or } B) = 0.50 \times 0.65 = 0.325

Event

C

is independent of both

A

and

B

, which implies

C

is independent of

(A \text{ or } B)

. Therefore, the joint probability is found by multiplying their individual probabilities.

Anahtar Kavram

Probability rules for mutually exclusive events (addition rule) and independent events (multiplication rule).

Soru 170Soru →

The continuous operating times of a model of industrial drone batteries are normally distributed with a mean of $220$ minutes and a standard deviation of $15$ minutes. A battery is designated as "high-efficiency" if its operating time places it in the top $16\%$ of all tested batteries. Based on the $68\text{--}95\text{--}99.7$ empirical rule for normal distributions, what is the minimum operating time, in minutes, required for a battery to be designated as high-efficiency?

205

220

235

236

250

Cevabı ve açıklamayı göster

Cevap: $235$

Cevap

235

minutes

According to the empirical rule for normal distributions,

68\%

of all observations fall within

1

standard deviation of the mean (

220 \pm 15

, or between

205

and

235

). Because normal distributions are symmetric, the remaining

32\%

of observations are split evenly between the upper and lower tails (

16\%

in each tail). The upper tail containing the top

16\%

of battery operating times starts at

1

standard deviation above the mean, which is

220 + 15 = 235

minutes.

Adım Adım Çözüm

1

Identify the given parameters of the normal distribution.

Mean

\mu = 220

minutes and standard deviation

\sigma = 15

minutes.

The problem specifies a normal distribution defined by these two parameters.

2

Apply the

68\text{--}95\text{--}99.7

empirical rule to determine the percentile threshold.

Approximately

68\%

of the distribution falls within

[\mu - \sigma, \mu + \sigma]

. The unshaded area (

100\% - 68\% = 32\%

) is split symmetrically, with

16\%

below

\mu - \sigma

and

16\%

above

\mu + \sigma

.

By symmetry of the normal curve, the top

16\%

corresponds precisely to values at or above

1

standard deviation above the mean (

z = +1

).

3

Calculate the raw score corresponding to

z = +1

.

Operating time

= \mu + 1\sigma = 220 + 1(15) = 235

minutes.

Adding one standard deviation to the mean yields the minimum score required to be in the upper tail containing

16\%

of the population.

Anahtar Kavram

Empirical Rule (68-95-99.7 Rule) and Symmetry of Normal Distributions

Soru 171Soru →

Two software security tools, Tool X and Tool Y, operate independently to scan code repositories for vulnerabilities. The probability that Tool X detects a specific type of security flaw is $0.80$ , and the probability that Tool Y detects the same flaw is $0.75$ . What is the probability that exactly one of the two tools detects the flaw?

Cevabı ve açıklamayı göster

Cevap: 0.35

Cevap

The probability that exactly one of the two tools detects the flaw is

0.35

.

The scenario requires finding the probability that exactly one tool detects the flaw. For independent events

X

and

Y

, 'exactly one' consists of two mutually exclusive events: (1) Tool X succeeds while Tool Y fails, which has probability

0.80 \times (1 - 0.75) = 0.80 \times 0.25 = 0.20

, and (2) Tool Y succeeds while Tool X fails, which has probability

0.75 \times (1 - 0.80) = 0.75 \times 0.20 = 0.15

. Summing these mutually exclusive probabilities gives

0.20 + 0.15 = 0.35

. Alternatively, one can subtract the probability of both tools succeeding (

0.80 \times 0.75 = 0.60

) from the probability of at least one tool succeeding (

0.80 + 0.75 - 0.60 = 0.95

), yielding

0.95 - 0.60 = 0.35

.

Adım Adım Çözüm

1

Determine the complement probabilities for each independent tool failing to detect the flaw.

P(\text{Not X}) = 1 - 0.80 = 0.20

and

P(\text{Not Y}) = 1 - 0.75 = 0.25

.

The probability of an event not occurring is equal to 1 minus the probability that it occurs.

2

Calculate the joint probability of Tool X detecting the flaw and Tool Y failing to detect it.

P(\text{X and Not Y}) = 0.80 \times 0.25 = 0.20

.

Because the tools operate independently, the joint probability is the product of their individual probabilities.

3

Calculate the joint probability of Tool Y detecting the flaw and Tool X failing to detect it.

P(\text{Y and Not X}) = 0.75 \times 0.20 = 0.15

.

Tool independence allows multiplying the individual probabilities of detection and non-detection.

4

Sum the probabilities of the two mutually exclusive scenarios representing 'exactly one tool detects the flaw'.

P(\text{Exactly One}) = 0.20 + 0.15 = 0.35

.

The events 'X only' and 'Y only' cannot happen simultaneously, so their probabilities add directly.

Anahtar Kavram

Independence and Mutual Exclusivity Rules in Compound Probability

Soru 172Soru →

The distribution of scores on a graduate admissions examination is normally distributed with a mean of $540$ and a standard deviation of $35$ . An applicant scored $610$ on this examination. If $800$ applicants scored higher than this applicant, which of the following is closest to the total number of applicants who took the examination?

8,000

16,000

32,000

50,000

64,000

Cevabı ve açıklamayı göster

Cevap: $32,000$

Cevap

32,000

The z-score for a score of

610

is calculated as

z = (610 - 540) / 35 = 2.0

. By the 68-95-99.7 empirical rule for normal distributions,

95\%

of all scores lie within

2

standard deviations of the mean. Because the distribution is symmetric, the remaining

5\%

is split equally between the two tails, meaning

2.5\%

of scores lie above

z = 2.0

. Given that

800

applicants scored higher than

610

, we set

0.025 N = 800

, which gives total applicants

N = 32,000

.

Adım Adım Çözüm

1

Calculate the z-score for a test score of

610

.

z = \frac{610 - 540}{35} = \frac{70}{35} = 2.0

The z-score measures how many standard deviations the score is above the mean.

2

Determine the proportion of scores that lie above

z = 2.0

using the 68-95-99.7 empirical rule.

Proportion

= \frac{100\% - 95\%}{2} = 2.5\% = 0.025

According to the empirical rule,

95\%

of values lie within

2

standard deviations of the mean (between

z = -2.0

and

z = 2.0

). Due to symmetry, half of the remaining

5\%

, or

2.5\%

, lies strictly above

z = 2.0

.

3

Set up an equation relating the number of higher-scoring applicants (

800

) to the total number of applicants (

N

).

0.025 \times N = 800 \implies N = \frac{800}{0.025} = 32,000

Since

2.5\%

of all applicants scored higher than

610

, dividing

800

by

0.025

yields the total population size.

Anahtar Kavram

Empirical Rule (68-95-99.7 Rule) and Standard Deviation Tail Areas

Tahmini Süre:1m 30s

Soru 173Soru →

The annual snowfall totals in a high-altitude meteorological district are normally distributed with a mean of $140$ inches and a standard deviation of $12$ inches. According to the 68–95–99.7 empirical rule for normal distributions, what percent of the years have an annual snowfall between $116$ inches and $152$ inches?

Cevabı ve açıklamayı göster

Cevap: 81.5

Cevap

81.5%

To find the percentage of data between

116

inches and

152

inches, first calculate the standard deviation distances (

z

-scores) from the mean of

140

inches.

116

inches is

24

inches below the mean, which corresponds to

z = -2

.

152

inches is

12

inches above the mean, which corresponds to

z = +1

. According to the 68–95–99.7 empirical rule,

95\%

of the data lies within

2

standard deviations of the mean, meaning

47.5\%

lies between

z = -2

and

z = 0

. Similarly,

68\%

of the data lies within

1

standard deviation of the mean, meaning

34\%

lies between

z = 0

and

z = +1

. Summing these two symmetric halves gives

47.5\% + 34\% = 81.5\%

.

Adım Adım Çözüm

1

Convert the boundary values (

116

inches and

152

inches) into standard z-scores.

z_{lower} = \frac{116 - 140}{12} = -2

and

z_{upper} = \frac{152 - 140}{12} = +1

Standardizing raw values into z-scores allows the application of standard normal distribution properties.

2

Apply the 68–95–99.7 empirical rule to split the area relative to the mean (

z = 0

).

Area from

z = -2

to

z = 0

is

47.5\%

; Area from

z = 0

to

z = +1

is

34\%

.

The normal curve is symmetrical around the mean. Thus,

95\%

between

-2\sigma

and

+2\sigma

yields

47.5\%

below the mean, and

68\%

between

-1\sigma

and

+1\sigma

yields

34\%

above the mean.

3

Sum the percentages of the two disjoint regions bounded by

z = -2

and

z = +1

.

47.5\% + 34\% = 81.5\%

Combining the area below the mean and the area above the mean gives the total percentage of values falling within the specified interval.

Anahtar Kavram

Normal distribution empirical rule (68–95–99.7 rule) with asymmetric standard deviation boundaries

Tahmini Süre:1m 30s

Soru 174Soru →

The pie chart below shows the distribution of a regional health network's total annual operating budget of $\$50,000,000$ among four divisions in 2025: Inpatient Care ( $40\%$ ), Outpatient Care ( $25\%$ ), Surgical Services ( $20\%$ ), and Medical Research ( $15\%$ ). The accompanying table details the breakdown of expenditures within the Medical Research division into three categories: Personnel ( $50\%$ ), Equipment ( $30\%$ ), and Operational Overhead ( $20\%$ ). How much greater is the total dollar amount allocated to Surgical Services than the combined dollar amount allocated to Equipment and Operational Overhead within Medical Research?

$3,750,000

$6,250,000

$7,750,000

$8,750,000

$15,000,000

Cevabı ve açıklamayı göster

Cevap: $6,250,000

Cevap

The total dollar amount allocated to Surgical Services is $6,250,000 greater than the combined allocation for Equipment and Operational Overhead within Medical Research.

Surgical Services receives

20\%

of the total

\$50,000,000

budget (

\$10,000,000

). Medical Research receives

15\%

of

\$50,000,000

(

\$7,500,000

). Equipment (

30\%

) and Operational Overhead (

20\%

) within Medical Research sum to

50\%

of the Medical Research budget, which equals

0.50 \times \$7,500,000 = \$3,750,000

. The difference is

\$10,000,000 - \$3,750,000 = \$6,250,000

.

Adım Adım Çözüm

1

Calculate the total dollar amount allocated to Surgical Services.

Surgical Services budget =

20\% \times \$50,000,000 = \$10,000,000

.

Surgical Services constitutes 20 percent of the network's overall budget.

2

Calculate the total dollar amount allocated to the Medical Research division.

Medical Research budget =

15\% \times \$50,000,000 = \$7,500,000

.

Medical Research constitutes 15 percent of the overall budget.

3

Determine the percentage and dollar amount for Equipment and Operational Overhead combined within Medical Research.

Combined percentage =

30\% + 20\% = 50\%

. Dollar amount =

50\% \times \$7,500,000 = \$3,750,000

.

The sub-table shows Equipment is 30 percent and Overhead is 20 percent of the Medical Research division's total budget.

4

Calculate the difference between the Surgical Services budget and the combined Research Equipment/Overhead budget.

Difference =

\$10,000,000 - \$3,750,000 = \$6,250,000

.

Subtracting the sub-category expenditures from the main division budget gives the required dollar difference.

Anahtar Kavram

Multi-tier data interpretation involving pie chart sector calculations combined with sub-table nested percentage breakdowns.

Data Analysis

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