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Romashka-Z-Leto [24]
3 years ago
9

A population of values has a normal distribution with μ = 158.9 μ = 158.9 and σ = 90.4 σ = 90.4 . You intend to draw a random sa

mple of size n = 218 n = 218 . Find P39, which is the mean separating the bottom 39% means from the top 61% means. P39 (for sample means) =
Mathematics
1 answer:
rosijanka [135]3 years ago
7 0

Answer:

P_{39}=133.68    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 158.9

Standard Deviation, σ = 90.4

We are given that the distribution is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.39

P( X < x) = P( z < \displaystyle\frac{x - 158.9}{90.4})=0.39  

Calculation the value from standard normal z table, we have,  

\displaystyle\frac{x - 158.9}{90.4} = -0.279\\\\x = 133.6784\approx 133.68  

P_{39}=133.68

133.68 separates the bottom 39% means from the top 61% means.

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Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t
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and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

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(a) <u>Probability</u> of distance at most <u>100m</u>, with λ = 0.0143:

F(100) = 1 - e^{-0.0143.100}

F(100) = 0.76

<u>Probability</u> of distance at most <u>200</u>:

F(200) = 1 - e^{-0.0143.200}

F(200) = 0.94

<u>Probability</u> of distance between <u>100 and 200</u>:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = \frac{1}{\lambda}

E(X) = \frac{1}{0.0143}

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = \sqrt{\frac{1}{\lambda^{2}} }

σ = \sqrt{\frac{1}{0.0143^{2}} }

σ = 69.93

<u>Distance exceeds the mean distance by more than 2σ</u>:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - e^{-0.0143*209.79})

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

\int\limits^m_0 f({x}) \, dx = 0.5

\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx = 0.5

\lambda.\frac{e^{-\lambda.x}}{-\lambda} = -e^{-\lambda.x} + e^{0}

1 - e^{-\lambda.m} = 0.5

-e^{-\lambda.m} = - 0.5

ln(e^{-0.0143.m}) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0
3 years ago
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