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2 years ago

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Dover Motors is a car dealership that sells new and used cars. Suppose they sold 140 used cars during the first quarter of 2011.

The average selling price was $10,325 with a standard deviation of $2,880. A random sample of 60 used cars from this population was selected. What is the probability that the sample mean exceeds $10,000?

1 answer:

Shalnov [3]2 years ago

5 0

Answer:

The probability that the sample mean exceeds $10,000 is 0.8078.

Step-by-step explanation:

The objective of the problem is obtained below:

From the given information, let X denotes the number of used cars which follows normal distribution with mean 10,325 and the standard deviation of 2,880 and sample size is 60 used cars. That is, u=10,325,δ = 2,880 and n=60

are used to estimate the probability that the sample mean exceeds 10,000.

See attached pictures.

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Answer:

a) Mean = 1030; Standard deviation = 12.38.

b) The county result is unusually high.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.

This first question is a binomial propability distribution.

We have a sample of 1210 Amricans, so $n = 1210$ .

The mean of the sample is 1030.

The probability of a success is $\pi = \frac{1030}{1210} = 0.8512$ .

The standard deviation of the sample is $s = \sqrt{n\pi(1-\pi)} = \sqrt{1210*0.8512*0.1488} = 12.38$

(b) Is that county result of 1030 unusually high, or low, or neither?

The first step is find the zscore when $X = 1030$ .

Then we find the pvalue of this zscore.

If this pvalue is bigger than 0.95, the county result is unusually high.

If this pvalue is smaller than 0.05, the county result is unusually low.

Otherwise, it is neither.

The national mean is 82%. So,

$\mu = 0.82(1210) = 992.2$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1030 - 992.2}{12.38}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989.This means that the county result is unusually high.

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Answer:

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Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose

Furkat [3]

Answer:

a) 0.164 = 16.4% probability that a disk has exactly one missing pulse

b) 0.017 = 1.7% probability that a disk has at least two missing pulses

c) 0.671 = 67.1% probability that neither contains a missing pulse

Step-by-step explanation:

To solve this question, we need to understand the Poisson distribution and the binomial distribution(for item c).

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

In which

x is the number of sucesses

$
e = 2.71828$ is the Euler number

$\mu$ is the mean in the given interval.

Binomial distribution:

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Poisson mean:

$\mu = 0.2$

a. What is the probability that a disk has exactly one missing pulse?

One disk, so Poisson.

This is P(X = 1).

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

0.164 = 16.4% probability that a disk has exactly one missing pulse

b. What is the probability that a disk has at least two missing pulses?

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

In which

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}
$

$P(X = 0) = \frac{e^{-0.2}*0.2^{0}}{(0)!} = 0.819$

$P(X = 1) = \frac{e^{-0.2}*0.2^{1}}{(1)!} = 0.164
$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.819 + 0.164 = 0.983$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.983 = 0.017$

0.017 = 1.7% probability that a disk has at least two missing pulses

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

Two disks, so binomial with n = 2.

A disk has a 0.819 probability of containing no missing pulse, and a 1 - 0.819 = 0.181 probability of containing a missing pulse, so $p = 0.181$

We want to find P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.181)^{0}.(0.819)^{2} = 0.671$

0.671 = 67.1% probability that neither contains a missing pulse

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2 years ago