Chris and Ben walked home from school. The distance Chris walked,in miles, is represented by point C on the number line. Ben wal
1 answer:
You might be interested in
Answer:
The answer to the questions are;
a. The probability that exactly six are retired people is 0.1633459.
b. The probability that 9 or more are retired people is 0.04677.
c. The number of expected retired people in a random sample of 25 stock investors is 0.179705.
d. In a random sample of 20 U.S. adults the probability that exactly eight adults invested in mutual funds is 0.179705.
e. The probability that fewer than five adults invested in mutual funds out of a random sample of 20 U.S. adults is 5.095×10⁻².
f. The probability that exactly one adult invested in mutual funds out of a random sample of 20 U.S. adults is 4.87×10⁻⁴.
g. The probability that 13 or more adults out of a random sample of 20 U.S. adults invested in mutual funds is 2.103×10⁻².
h. 4, 1, 13. They tend to converge to the probability of the expected value.
Step-by-step explanation:
To solve the question, we note that the binomial distribution probability mass function is given by
f(n,p,x) = × pˣ × (1-p)ⁿ⁻ˣ = ₙCₓ × pˣ × (1-p)ⁿ⁻ˣ
Also the mean of the Binomial distribution is given by
Mean = μ = n·p = 25 × 0.2 = 5
Variance = σ² = n·p·(1-p) = 25 × 0.2 × (1-0.2) = 4
Standard Deviation = σ =
Since the variance < 5 the normal distribution approximation is not appropriate to sole the question
We proceed as follows
a. The probability that exactly six are retired people is given by
f(25, 0.2, 6) = ₂₅C₆ × 0.2⁶ × (1-0.2)¹⁹ = 0.1633459.
b. The probability that 9 or more are retired people is given by
P(x>9) = 1- P(x≤8) = 1- ∑f(25, 0.2, x where x = 0 →8)
Therefore we have
f(25, 0.2, 0) = ₂₅C₀ × 0.2⁰ × (1-0.2)²⁵ = 3.78×10⁻³
f(25, 0.2, 1) = ₂₅C₁ × 0.2¹ × (1-0.2)²⁴ = 2.36 ×10⁻²
f(25, 0.2, 2) = ₂₅C₂ × 0.2² × (1-0.2)²³ = 7.08×10⁻²
f(25, 0.2, 3) = ₂₅C₃ × 0.2³ × (1-0.2)²² = 0.135768
f(25, 0.2, 4) = ₂₅C₄ × 0.2⁴ × (1-0.2)²¹ = 0.1866811
f(25, 0.2, 5) = ₂₅C₅ × 0.2⁵ × (1-0.2)²⁰ = 0.1960151
f(25, 0.2, 6) = ₂₅C₆ × 0.2⁶ × (1-0.2)¹⁹ = 0.1633459
f(25, 0.2, 7) = ₂₅C₇ × 0.2⁷ × (1-0.2)¹⁸ = 0.11084187
f(25, 0.2, 8) = ₂₅C₈ × 0.2⁸ × (1-0.2)¹⁷ = 6.235×10⁻²
∑f(25, 0.2, x where x = 0 →8) = 0.953226
and P(x>9) = 1- P(x≤8) = 1 - 0.953226 = 0.04677.
c. The number of expected retired people in a random sample of 25 stock investors is given by
Proportion of retired stock investors × Sample count
= 0.2 × 25 = 5.
d. In a random sample of 20 U.S. adults the probability that exactly eight adults invested in mutual funds is given by
Here we have p = 0.4 and n·p = 8 while n·p·q = 4.8 which is < 5 so we have
f(20, 0.4, 8) = ₂₀C₈ × 0.4⁸ × (1-0.4)¹² = 0.179705.
e. The probability that fewer than five adults invested in mutual funds out of a random sample of 20 U.S. adults is
P(x<5) = ∑f(20, 0.4, x, where x = 0 →4)
Which gives
f(20, 0.4, 0) = ₂₀C₀ × 0.4⁰ × (1-0.4)²⁰ = 3.66×10⁻⁵
f(20, 0.4, 1) = ₂₀C₁ × 0.4¹ × (1-0.4)¹⁹ = 4.87×10⁻⁴
f(20, 0.4, 2) = ₂₀C₂ × 0.4² × (1-0.4)¹⁸ = 3.09×10⁻³
f(20, 0.4, 3) = ₂₀C₃ × 0.4³ × (1-0.4)¹⁷ = 1.235×10⁻²
f(20, 0.4, 4) = ₂₀C₄ × 0.4⁴ × (1-0.4)¹⁶ = 3.499×10⁻²
Therefore P(x<5) = 5.095×10⁻².
f. The probability that exactly one adult invested in mutual funds out of a random sample of 20 U.S. adults is given by
f(20, 0.4, 1) = ₂₀C₁ × 0.2¹ × (1-0.2)¹⁹ = 4.87×10⁻⁴.
g. The probability that 13 or more adults out of a random sample of 20 U.S. adults invested in mutual funds is
P(x≥13) = ∑f(20, 0.4, x where x = 13 →20) we have
f(20, 0.4, 13) = ₂₀C₁₃ × 0.4¹³ × (1-0.4)⁷ = 1.46×10⁻²
f(20, 0.4, 14) = ₂₀C₁₄ × 0.4¹⁴ × (1-0.4)⁶ = 4.85×10⁻³
f(20, 0.4, 15) = ₂₀C₁₅ × 0.4¹⁵ × (1-0.4)⁵ = 1.29×10⁻³
f(20, 0.4, 16) = ₂₀C₁₆ × 0.4¹⁶ × (1-0.4)⁴ = 2.697×10⁻⁴
f(20, 0.4, 17) = ₂₀C₁₇ × 0.4¹⁷ × (1-0.4)³ = 4.23×10⁻⁵
f(20, 0.4, 18) = ₂₀C₁₈ × 0.4¹⁸ × (1-0.4)² = 4.70×10⁻⁶
f(20, 0.4, 19) = ₂₀C₁₉ × 0.4¹⁹ × (1-0.4)⁴ = 3.299×10⁻⁷
f(20, 0.4, 20) = ₂₀C₂₀ × 0.4²⁰ × (1-0.4)⁰ = 1.0995×10⁻⁸
P(x≥13) = 2.103×10⁻².
h. For part e we have exactly 4 with a probability of 3.499×10⁻²
For part f the probability for the one adult is 4.87×10⁻⁴
For part g, we have exactly 13 with a probability of 1.46×10⁻²
The expected number is 8 towards which the exact numbers with the highest probabilities in parts e to g are converging.