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%.

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%.

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%.

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:

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%.

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%.

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%.

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%.

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%.

9.
If $p$ = 0.50 and $n$ = 100, what is the value of μ$\stackrel{ˆ}{p}$, σ $\stackrel{ˆ}{p}$?
 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.

10.
Suppose $n$ = 50 and $p$ = 0.35. Can we safely approximate $\stackrel{ˆ}{p}$ 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.