Ask question

Ask question

All categories

3 years ago

10

For the following exercise, determine the range (possible values) of the random variable, X. A batch of 300 machined parts conta

ins 10 that do not conform to customer requirements. The random variable is the number of parts in a sample of 5 parts that do not conform to customer requirements. Choose the correct answer.
a. The range of X is {0,1,2,3,4,5,6}
b. The range of X is {0,1,2,3,4,5,6}
c. The range of X is {0,1,2,3,4,5,6}
d. The range of X is {0,1,2,3,4,5,6}
e. The range of X is {0,1,2,3,4,5,6}

1 answer:

solmaris [256]3 years ago

5 0

Answer:

{0,1,2,3,4,5}

Step-by-step explanation:

We are given that

Total number of machine parts=300

Number of defective machine parts=10

Total number of good machine parts=300-10=290

Sample contain parts that do not conform to customer requirement=5

X is a random variable which is the number of parts in a sample of 5 parts that do not conform to customer requirements.

We have to find the correct answer.

The sample contain 5 parts

Therefore, the possible values of random variable X

0,1,2,3,4,5

Hence, the range of X is given by

{0,1,2,3,4,5}

You might be interested in

3. Using the sample size of 800 and the proportion 0.47, calculate the margin of error associated with the estimate of the propo

salantis [7]

Answer:

$ME=1.96\sqrt{\frac{0.47 (1-0.47)}{800}}=0.0346$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

We assume for this case a confidence level of 95%. In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And if we replace the values obtained we got this:

$ME=1.96\sqrt{\frac{0.47 (1-0.47)}{800}}=0.0346$

6 0

3 years ago

Can someone correct me I don't know if I got it right

Angelina_Jolie [31]

Answer:

yup

Step-by-step explanation:

8 0

3 years ago

Choose the correct correspondence RT

Marizza181 [45]

<u></u> $RT$ corresponds to TR. correct option b.

<u>Step-by-step explanation:</u>

In the given parallelogram or rectangle , we have a diagonal RT . We need to find which side is in correspondence with side/Diagonal RT of parallelogram URST .

<u>Side TU:</u>

In triangle UTR , we see that TR is hypotenuse and is the longest side among UR & TU . So , TR can never be equal in length to UR & TU . So there's no correspondence of Side TU with RT.

<u>Side TR:</u>

Since, direction of sides are not mentioned here , we can say that TR & RT is parallel & equal to each other . So , TR is in correspondence with side/Diagonal RT of parallelogram URST .

<u>Side UR:</u>

In triangle UTR , we see that TR is hypotenuse and is the longest side among UR & TU . So , TR can never be equal in length to UR & TU . So there's no correspondence of Side UR with RT.

4 0

3 years ago

Find the earnings for giving a haircut to m men and w women. Justify your answer

Ratling [72]

Woman and men would be the same price, if the haircuts are very popular, each gender got the same haircut, then they also would be the same price, it is mostly about the prices, interestingness, and popularity or each barbershop, go by your nearest barbershop and see what prices they have. :))

3 0

3 years ago

A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is

scoray [572]

Answer:

Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

$s_{1}=50000$ represent the sample standard deviation for downtown

$s_{2}=40000$ represent the sample standard deviation for the freeway

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

Null hypothesis: $\mu_{1}\leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value.

Based on the significance level $\alpha=0.05$ and $\alpha/2=0.025$ we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is $t_{1-\alpha/2}=1.66$

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

$t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874$

What is the p-value for this hypothesis test?

Since is a right tailed test the p value would be:

$p_v =P(t_{70}>1.874)=0.033$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.

The best option would be:

Yes, since the test statistic value is greater than the critical value.

5 0

3 years ago