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2 years ago

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Suppose you take a simple random sample of size n=175 from the population of Reese's Pieces, still assuming that 45% of this pop

ulation is orange (LaTeX: \piπ=.45).
What is the probability that a sample proportion of orange candies will be between .35 and .55?

1 answer:

Gnesinka [82]2 years ago

3 0

Answer:

The probability is $P(0.35$

Step-by-step explanation:

From the question we are told that

The sample size is n = 175

The population proportion is p = 0.45

Generally the mean of the sampling distribution is $\mu_{x} = 0.45$

Generally the standard deviation is mathematically represented as

$\sigma_{x} = \sqrt{\frac{p (1-p)}{n} }$

=> $\sigma_{x} = \sqrt{\frac{0.45 (1-0.45)}{175} }$

=> $\sigma_{x} = 0.0376$

Generally the probability of that the sample proportion of orange candies will be between 0.35 and 0.55 is

$P(0.35 < X < 0.55) = P( \frac{0.35 - 0.45}{0.0376} < \frac{X -\mu_{x}}{\sigma_{x}} < \frac{0.55 - 0.45}{0.0376} )$

=> $P(0.35$

Generally $\frac{X - \mu_{x}}{\sigma_{x}} = Z (The \ standardized \ value \ of X)$

So

$P(0.35$

=> $P(0.35$

From the z-table

$P( Z< 2.6595 ) = 0.99609$

and

$P( Z< - 2.6595 ) = 0.0039128$

So

$P(0.35$

=> $P(0.35$

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The direction cosines are:

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The direction angles are:

40°, 81° and 52° with respect to the x, y and z axes respectively.

Step-by-step explanation:

For a given vector a = ai + aj + ak, its direction cosines are the cosines of the angles which it makes with the x, y and z axes.

If a makes angles α, β, and γ (which are the direction angles) with the x, y and z axes respectively, then its direction cosines are: cos α, cos β and cos γ in the x, y and z axes respectively.

Where;

cos α = $\frac{a . i}{|a| . |i|}$ ---------------------(i)

cos β = $\frac{a.j}{|a||j|}$ ---------------------(ii)

cos γ = $\frac{a.k}{|a|.|k|}$ ----------------------(iii)

<em>And from these we can get the direction angles as follows;</em>

α = cos⁻¹ ( $\frac{a . i}{|a| . |i|}$ )

β = cos⁻¹ ( $\frac{a.j}{|a||j|}$ )

γ = cos⁻¹ ( $\frac{a.k}{|a|.|k|}$ )

Now to the question:

Let the given vector be

a = 5i + j + 4k

a . i = (5i + j + 4k) . (i)

a . i = 5 [a.i <em>is just the x component of the vector</em>]

a . j = 1 [<em>the y component of the vector</em>]

a . k = 4 [<em>the z component of the vector</em>]

<em>Also</em>

|a|. |i| = |a|. |j| = |a|. |k| = |a| [since |i| = |j| = |k| = 1]

|a| = $\sqrt{5^2 + 1^2 + 4^2}$

|a| = $\sqrt{25 + 1 + 16}$

|a| = $\sqrt{42}$

Now substitute these values into equations (i) - (iii) to get the direction cosines. i.e

cos α = $\frac{5}{\sqrt{42} }$

cos β = $\frac{1}{\sqrt{42} }$

cos γ = $\frac{4}{\sqrt{42} }$

From the value, now find the direction angles as follows;

α = cos⁻¹ ( $\frac{a . i}{|a| . |i|}$ )

α = cos⁻¹ ( $\frac{5}{\sqrt{42} }$ )

α = cos⁻¹ ( $\frac{5}{6.481}$ )

α = cos⁻¹ (0.7715)

α = 39.51

α = 40°

β = cos⁻¹ ( $\frac{a.j}{|a||j|}$ )

β = cos⁻¹ ( $\frac{1}{\sqrt{42} }$ )

β = cos⁻¹ ( $\frac{1}{6.481 }$ )

β = cos⁻¹ ( 0.1543 )

β = 81.12

β = 81°

γ = cos⁻¹ ( $\frac{a.k}{|a|.|k|}$ )

γ = cos⁻¹ ( $\frac{4}{\sqrt{42} }$ )

γ = cos⁻¹ ( $\frac{4}{6.481}$ )

γ = cos⁻¹ (0.6172)

γ = 51.89

γ = 52°

<u>Conclusion:</u>

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$\frac{5}{\sqrt{42} }$ , $\frac{1}{\sqrt{42} }$ and $\frac{4}{\sqrt{42} }$ with respect to the x, y and z axes respectively.

The direction angles are:

40°, 81° and 52° with respect to the x, y and z axes respectively.

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