If 3x + 7 = 19, find the value of 4x + 1.<br> 4x + 1 =

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard

avanturin [10]

Answer:

There is a 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 33,208 miles, with a standard deviation of 2503 miles.

This means that $\mu = 33208, \sigma = 2503$ .

What is the probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct?

This is the pvalue of Z when $X = 33208+633 = 33841$ subtracted by the pvalue of Z when $X = 33208 - 633 = 32575$

By the Central Limit Theorem, we have t find the standard deviation of the sample, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{2503}{\sqrt{49}} = 357.57$

So

X = 33841

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

$Z = \frac{33841 - 33208}{357.57}$

$Z = 1.77$

$Z = 1.77$ has a pvalue of 0.9616

X = 32575

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

$Z = \frac{32575- 33208}{357.57}$

$Z = -1.77$

$Z = -1.77$ has a pvalue of 0.0384.

This means that there is a 0.9616 - 0.0384 = 0.9232 = 92.32% probability that the sample mean would differ from the population mean by less than 633 miles in a sample of 49 tires if the manager is correct.

4 0

3 years ago

Which expression has the same value as 15+24

lakkis [162]

15+24= put it in the calculator u will find the answer(hope this helped)

6 0

3 years ago

Read 2 more answers

You have 5/6 cup of cocoa in your cupboard. A recipe for cookies calls for 2/5 cup of cocoa. How much cocoa would you have left

bixtya [17]

5/6-2/5= 13/30 .........

7 0

3 years ago

-4 is less than or equal to 2t + 5 which is less than 11

dlinn [17]

Not sure what you’re trying to find but here’s the simplified version:

-4<=2t+5<11
-9<=2t<6
-9/2<=t<3

7 0

3 years ago

Read 2 more answers

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

If 3x + 7 = 19, find the value of 4x + 1. 4x + 1 =