Mathematics, 22.08.2020 20:01 mjwaple57

Forty percent of all Americans who travel by car look for gas stations and food outlets that are close to or visible from the highway. Suppose a random sample of n=25 Americans who travel by car are asked how they determine where to stop for food and gas. Let x be the number in the sample who respond that they look for gas stations and food outlets that are close to or visible from the highway. a. What are the mean and variance of x?
b. Calculate the interval μ±2σμ±2σ. What values of the binomial random variable x fall into this interval?
c. Find P(6≤≤x$\leq$14). How does this compare with the fraction in the interval μ±2σμ±2σ for any distribution? For mound-shaped distributions?

Answers: 1

Show answers

Answers

Answer from: seonna

Probabilities are used to determine the chance of an event.

$\mathbf{Mean = 10}$ and $\mathbf{Variance = 6}$ .97.72% values of the binomial random variable x fall into the interval $\mathbf{\mu \pm 2\sigma}$ .93.61% values of the binomial random variable x fall into the interval 6 to 14

The given parameters are:

$\mathbf{n = 25}$

$\mathbf{p = 40\%}$

(a) Mean and variance

The mean is calculated as follows:

$\mathbf{Mean = np}$

$\mathbf{Mean = 25 \times 40\%}$

$\mathbf{Mean = 10}$

The variance is calculated as follows:

$\mathbf{Variance = np(1 - p)}$

So, we have:

$\mathbf{Variance = 25 \times 40\%(1 - 40\%)}$

$\mathbf{Variance = 6}$

(b) The interval $\mathbf{\mu \pm 2\sigma}$

First, we calculate the standard deviation

$\mathbf{\sigma = \sqrt{Variance}}$

$\mathbf{\sigma = \sqrt{6}}$

$\mathbf{\sigma = 2.45}$

So, we have:

$\mathbf{\mu \pm 2\sigma = 10 \pm 2 \times 2.45}$

$\mathbf{\mu \pm 2\sigma = 10 \pm 4.90}$

Split

$\mathbf{\mu \pm 2\sigma = 10 + 4.90\ or\ 10 - 4.90}$

$\mathbf{\mu \pm 2\sigma = 14.90\ or\ 5.10}$

Approximate

$\mathbf{\mu \pm 2\sigma = 15\ or\ 5}$

So, we have:

$\mathbf{\mu \pm 2\sigma = (5,15)}$

The binomial probability is then calculated as:

$\mathbf{P = ^nC_x p^x \times (1 - p)^{n - x}}$

This gives

$\mathbf{P = ^{25}C_5 \times (0.4)^5 \times (1 - 0.6)^{25 - 5} + ...... +^{25}C_{15} \times (0.4)^{15} \times (1 - 0.6)^{25 - 15}}$

$\mathbf{P = 0.0199 + ..... + 0.0434 + 0.0212}$

$\mathbf{P = 0.9772}$

Express as percentage

$\mathbf{P = 97.72\%}$

This means that; 97.72% values of the binomial random variable x fall into the interval $\mathbf{\mu \pm 2\sigma}$

$\mathbf{(c)\ P(6 \le x \le 14)}$

The binomial probability is then calculated as:

$\mathbf{P = ^nC_x p^x \times (1 - p)^{n - x}}$

So, we have:

$\mathbf{P = ^{25}C_6 \times (0.4)^6 \times (1 - 0.4)^{25 - 6} + ...... +^{25}C_{14} \times (0.4)^{14} \times (1 - 0.4)^{25 - 14}}$

$\mathbf{P = 0.0422 +.............+0.0759 + 0.0434}$

$\mathbf{P = 0.9361}$

This means that:

93.61% values of the binomial random variable x fall into the interval 6 to 14

By comparison, 93.61% is very large compared to the other distributions., and the proportion 93.61 is slightly less when compared to the mound-shaped distribution.

Read more about binomial probability at:

link

Answer from: pandamaknae2003

Explained below.

Step-by-step explanation:

Let the random variable X be defined as the number of Americans who travel by car look for gas stations and food outlets that are close to or visible from the highway.

The probability of the random variable X is: p = 0.40.

A random sample of n =25 Americans who travel by car are selected.

The events are independent of each other, since not everybody look for gas stations and food outlets that are close to or visible from the highway.

The random variable X follows a Binomial distribution with parameters n = 25 and p = 0.40.

(a)

The mean and variance of X are:

$\mu=np=25\times 0.40=10\\\\\sigma^{2}=np(1-p)-25\times0.40\times(1-0.40)=6$

Thus, the mean and variance of X are 10 and 6 respectively.

(b)

Compute the values of the interval μ ± 2σ as follows:

$\mu\pm 2\sigma=(\mu-2\sigma, \mu+ 2\sigma)$

$=(10-2\cdot\sqrt{6},\ 10+2\cdot\sqrt{6})\\\\=(5.101, 14.899)\\\\\approx (5, 15)$

Compute the probability of P (5 ≤ X ≤ 15) as follows:

$P(5\leq X\leq 15)=\sum\limits^{15}_{x=5}{{25\choose x}(0.40)^{x}(1-0.40)^{25-x}}$

$=0.0199+0.0442+0.0799+0.1199+0.1511+0.1612\\+0.1465+0.1140+0.0759+0.0434+0.0212\\\\=0.9772$

Thus, 97.72% values of the binomial random variable x fall into this interval.

(c)

Compute the value of P (6 ≤ X ≤ 14) as follows:

$P(6\leq X\leq 14)=\sum\limits^{14}_{x=6}{{25\choose x}(0.40)^{x}(1-0.40)^{25-x}}$

$=0.0442+0.0799+0.1199+0.1511+0.1612\\+0.1465+0.1140+0.0759+0.0434\\\\=0.9361\\\\\approx P(5\leq X\leq 15)$

The value of P (6 ≤ X ≤ 14) is 0.9361.

According to the Tchebysheff's theorem, for any distribution 75% of the data falls within μ ± 2σ values.

The proportion 0.9361 is very large compared to the other distributions.

Whereas for a mound-shaped distributions, 95% of the data falls within μ ± 2σ values. The proportion 0.9361 is slightly less when compared to the mound-shaped distribution.

Answer from: Quest

1+1=22×1=22-1=11-1=1duh

Answer from: Quest

sample answer: you can determine whether each element of the domain is paired with exactly one element of the range. for example, if given a graph, you could use the vertical line test; if a vertical line intersects the graph more than once, then the relation that the graph represents is not a function.