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Standard Deviation Word Problems

Standard Deviation Word Problems
  • Page 1
 1.  
Find the probability that a student can guess 60% or less of the answers correctly in a 40 question true false examination.
a.
92.27%
b.
13.41%
c.
36.59%
d.
86.59%


Solution:

Probability of answering a question correctly, p = 1 / 2.

q = 1 - p = 1 - 0.5 = 0.5
n = 40.

For a random variable pˆ = rn to be approximated by a normal random variable with mean, μ pˆ, and standard deviation, σ pˆ, it has to satisfy the following conditions:
np > 5 and nq > 5

np = 40 × 0.5 = 20 > 5
nq = 40 × 0.5 = 20 > 5
Since both the conditions are satisfied, the random variable pˆ can be approximated by a normal random variable.

Mean μ pˆ = p = 0.50.

Standard deviation σ pˆ = pqn = 0.50 × 0.5040 = 0.079.

Continuity correction = 0.5n = 0.540 = 0.0125.

To convert the pˆ value distribution to x distribution, we add 0.0125 to the given value.

P(pˆ ≤ 0.60) P(x ≤ 0.6125)

                            P(z ≤ 0.6125-0.50.079)

P(z ≤ 1.424)

0.9227.

The probability that a student can guess 60% or more of the answers correctly in a 40 question true false examination is 92.27%.


Correct answer : (1)
 2.  
It has been found that 6% of the tools produced by a certain machine are defective. What is the probability that in a shipment of 500 such tools 6% or more will prove defective?
a.
6%
b.
53.98%
c.
3.98%
d.
0.54%


Solution:

Probability that a tool produced is defective, p = 0.06.

q = 1 - p = 1 - 0.06 = 0.94
n = 500.

For a random variable pˆ = rn to be approximated by a normal random variable with mean, μ pˆ, and standard deviation, σ pˆ, it has to satisfy the following conditions:
np > 5 and nq > 5

np = 500 × 0.06 = 30 > 5
nq = 500 × 0.94 = 470 > 5
Since both the conditions are satisfied, the random variable pˆ can be approximated by a normal random variable.

Mean μ pˆ = p = 0.06.

Standard deviation σ pˆ = pqn = 0.06 × 0.94500 = 0.01.

Continuity correction = 0.5n = 0.5500 = 0.001

To convert the pˆ value distribution to x distribution, we subtract 0.001 from the given value.

P(pˆ ≥ 0.06) P(x ≥ 0.059)

                            P(z ≥ 0.059-0.060.01)

P(z ≥ -0.1)

0.5398.

The probability that in a shipment of 500 such tools 6% or more will prove defective is 53.98%.


Correct answer : (2)
 3.  
It has been found that 6% of the tools produced by a certain machine are defective. What is the probability that in a shipment of 500 such tools 4% or less will prove defective?
a.
97.13%
b.
0.97%
c.
2.87%
d.
47.13%


Solution:

Probability that a tool produced is defective, p = 0.06.

q = 1 - p = 1 - 0.06 = 0.94
n = 500.

For a random variable pˆ = rn to be approximated by a normal random variable with mean, μ pˆ, and standard deviation, σ pˆ, it has to satisfy the following conditions:
np > 5 and nq > 5

np = 500 × 0.06 = 30 > 5
nq = 500 × 0.94 = 470 > 5
Since both the conditions are satisfied, the random variable pˆ can be approximated by a normal random variable.

Mean μ pˆ = p = 0.06.

Standard deviation σ pˆ = pqn = 0.06 × 0.94500 = 0.01.

Continuity correction = 0.5n = 0.5500 = 0.001.

To convert the pˆ value distribution to x distribution, we add 0.001 to the given value.

P(pˆ ≤ 0.04) P(x ≤ 0.041)

                            P(z ≤ 0.041-0.060.01)

P(z ≤ -1.9)

0.0287.

The probability that in a shipment of 500 such tools 4% or less will prove defective is 2.87%.


Correct answer : (3)
 4.  
The election results showed that a certain candidate received 44% of the votes. Determine the probability that a poll of 500 people selected at random from the voting population would have shown a majority of votes in favor of the candidate.
a.
49.71%
b.
0.29%
c.
0.0029%
d.
99.71%


Solution:

Percentage of votes the candidate received p = 44% = 0.44

q = 1 - p = 1 - 0.44 = 0.56
n = 500

For a random variable pˆ = rn to be approximated by a normal random variable with mean, μ pˆ, and standard deviation, σ pˆ, it has to satisfy the following conditions:
np > 5 and nq > 5

np = 500 × 0.44 = 220 > 5
nq = 500 × 0.56 = 280 > 5
Since both the conditions are satisfied, the random variable pˆ can be approximated by a normal random variable.

Mean μ pˆ = p = 0.44

Standard deviation σ pˆ = pqn = 0.44 × 0.56500 = 0.022

Continuity correction = 0.5n = 0.5500 = 0.001

To get a majority, the candidate should get more than 50% of the votes. That is pˆ > 0.5

To convert the pˆ value distribution to x distribution, we add 0.001 to the given value.

P(pˆ > 0.5) P(x > 0.501)

                            P(z > 0.501-0.440.022)

P(z > 2.77)

0.0029

The probability that the voting population would have shown a majority of votes in favor of the candidate is 0.0029 or 0.29%.


Correct answer : (2)
 5.  
Find the probability that of the next 300 children born, 60% or greater will be boys. Assume equal probabilities for the births of boys and girls.
a.
50.04%
b.
99.96%
c.
49.96%
d.
0.04%


Solution:

Probability that a child born is a boy, p = 1 / 2 = 0.5
[Equal probabilities for the births of boys and girls.]

q = 1 - p = 1 - 0.5 = 0.5
n = 300

For a random variable pˆ = rn to be approximated by a normal random variable with mean, μ pˆ, and standard deviation, σ pˆ, it has to satisfy the following conditions:
np > 5 and nq > 5

np = 300 × 0.5 = 150 > 5
nq = 300 × 0.5 = 150 > 5
Since both the conditions are satisfied, the random variable pˆ can be approximated by a normal random variable.

Mean μ pˆ = p = 0.5

Standard deviation σ pˆ = pqn = 0.5 × 0.5300 = 0.029

Continuity correction = 0.5n = 0.5300 = 0.0017

To convert the pˆ value distribution to x distribution, we subtract 0.0017 from the given value.

P(pˆ ≥ 0.6) P(x ≥ 0.5983)

                            P(z ≥ 0.5983-0.50.029)

P(z ≥ 3.39)

0.0004

The probability that of the next 300 children born, 60% or greater will be boys is 0.04%.


Correct answer : (4)
 6.  
The probability that a student has access to Internet from his/her home is 40%. If a random sample of 100 students is selected, then find the probability that between 50% and 60% of the students have access to Internet from their home.
a.
2.61%
b.
0.026%
c.
97.39%
d.
26.1%


Solution:

Probability that a student has access to Internet from home p = 40% = 0.4

q = 1 - p = 1 - 0.4 = 0.6
n = 100

For a random variable pˆ = rn to be approximated by a normal random variable with mean, μ pˆ, and standard deviation, σ pˆ, it has to satisfy the following conditions:
np > 5 and nq > 5

np = 100 × 0.4 = 40 > 5
nq = 100 × 0.6 = 60 > 5
Since both the conditions are satisfied, the random variable pˆ can be approximated by a normal random variable.

Mean μ pˆ = p = 0.4

Standard deviation σ pˆ = pqn = 0.4 × 0.6100 = 0.049

Continuity correction = 0.5n = 0.5100 = 0.005

We subtract 0.005 from the left pˆ endpoint and add 0.005 to the right pˆ endpoint. The x interval is from 0.495 to 0.605.

P(0.5 ≤ pˆ ≤ 0.6) P(0.495 ≤ x ≤ 0.605)

                            P(0.495-0.40.049 ≤ z ≤ 0.605-0.40.049)

P(1.94 ≤ z ≤ 4.18)

0.4999 - 0.4738

0.0261

The probability that between 50% and 60% of the students have access to Internet from their home is 2.61%.


Correct answer : (1)
 7.  
Find the probability that of the next 100 children born, 40% or less will be boys. Assume equal probabilities for the births of boys and girls.
a.
1.78%
b.
98.22%
c.
2.87%
d.
97.13%


Solution:

Probability that a child born is a boy, p = 1 / 2 = 0.5
[Equal probabilities for the births of boys and girls.]

q = 1 - p = 1 - 0.5 = 0.5
n = 100

For a random variable pˆ = rn to be approximated by a normal random variable with mean, μ pˆ, and standard deviation, σ pˆ, it has to satisfy the following conditions:
np > 5 and nq > 5

np = 100 × 0.5 = 50 > 5
nq = 100 × 0.5 = 50 > 5
Since both the conditions are satisfied, the random variable pˆ can be approximated by a normal random variable.

Mean μ pˆ = p = 0.5

Standard deviation σ pˆ = pqn = 0.5 × 0.5100 = 0.05

Continuity correction = 0.5n = 0.5100 = 0.005

To convert the pˆ value distribution to x distribution, we add 0.005 to the given value.

P(pˆ ≤ 0.40) P(x ≤ 0.405)

                       P(z ≤ 0.405-0.50.05)

P(z ≤ -1.9)

0.0287

The probability that of the next 100 children born, 40% or less will be boys is 2.87%.


Correct answer : (3)
 8.  
Find the probability that of the next 200 children born, between 45% and 60% will be girls. Assume equal probabilities for the births of boys and girls.
a.
43.31%
b.
49.82%
c.
6.51%
d.
93.13%


Solution:

Probability that a child born is a girl, p = 1 / 2 = 0.5
[Equal probabilities for the births of boys and girls.]

q = 1 - p = 1 - 0.5 = 0.5
n = 200

For a random variable pˆ = rn to be approximated by a normal random variable with mean, μ pˆ, and standard deviation, σ pˆ, it has to satisfy the following conditions:
np > 5 and nq > 5

np = 200 × 0.5 = 100 > 5
nq = 200 × 0.5 = 100 > 5
Since both the conditions are satisfied, the random variable pˆ can be approximated by a normal random variable.

Mean μ pˆ = p = 0.5

Standard deviation σ pˆ = pqn = 0.5 × 0.5200 = 0.035

Continuity correction = 0.5n = 0.5200 = 0.0025

We subtract 0.0025 from the left pˆ endpoint and add 0.0025 to the right pˆ endpoint. The x interval is from 0.4475 to 0.6025.

P(0.45 ≤ pˆ ≤ 0.60) P(0.4475 ≤ x ≤ 0.6025)

                            P(0.4475-0.50.035 ≤ z ≤ 0.6025-0.50.035)

P(- 1.5 ≤ z ≤ 2.92)

0.4331 + 0.4982

0.9313

The probability that of the next 200 children born, between 45% and 60% will be girls is 93.13%.


Correct answer : (4)
 9.  
If p = 0.50 and n = 100, what is the value of μpˆ, σ pˆ?
a.
50.0, 5.0
b.
0.50, 0.05
c.
0.50, 5.0
d.
0.50, 0.50


Solution:

p = 0.50, n = 100

q = 1 - p = 1 - 0.5 = 0.5

For a random variable pˆ = rn to be approximated by a normal random variable with mean, μ pˆ, and standard deviation, σ pˆ, it has to satisfy the following conditions:
np > 5 and nq > 5

np = 100 × 0.5 = 50 > 5
nq = 100 × 0.5 = 50 > 5
Since both the conditions are satisfied, the random variable pˆ can be approximated by a normal random variable.

Mean μ pˆ = p = 0.50

Standard deviation σ pˆ = pqn = 0.50 × 0.50100 = 0.05

So, the mean and standard deviation are 0.50, 0.05 respectively.


Correct answer : (2)
 10.  
Suppose n = 50 and p = 0.35. Can we safely approximate pˆ by a normal distribution? What is the value of the continuity correction?
a.
Yes, 0.01
b.
Yes, 25
c.
Yes, 0.07
d.
No, 0.01


Solution:

p = 0.35, n = 50

q = 1 - p = 1 - 0.35 = 0.65

For a random variable pˆ = rn to be approximated by a normal random variable with mean, μ pˆ, and standard deviation, σ pˆ, it has to satisfy the following conditions:
np > 5 and nq > 5

np = 50 × 0.35 = 17.5 > 5
nq = 50 × 0.65 = 32.5 > 5
Since both the conditions are satisfied, the random variable pˆ can be approximated by a normal random variable.

Continuity correction = 0.5n = 0.550 = 0.01

So, the continuity correction is 0.01.


Correct answer : (1)

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