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3 years ago

10

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 15 years, and standard deviation of 3

.7 years. If you randomly purchase 22 items, what is the probability that their mean life will be longer than 13 years?

1 answer:

Gala2k [10]3 years ago

7 0

Answer:

The probability that their mean life will be longer than 13 years = .9943 or 99.43%

Step-by-step explanation:

Given -

Mean $(\nu )$ = 15 years

Standard deviation $(\sigma )$ = 3.7 years

Sample size ( n ) =22

Let $\overline{X}$ be the mean life of manufacturing items

the probability that their mean life will be longer than 13 years =

$P(\overline{X}> 13)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{13 - 15 }{\frac{3.7}{\sqrt{22}}})$ $(Z =\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}})$

= $P(Z> -2.53)$

= $1 - P(Z < -2.53)$ Using Z table

= 1 - .0057

= .9943

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Read 2 more answers

You go to a local mechanic to get your tires changed. The tires cost $300. There is a 6% sales tax, but you get a 10% discount.

Romashka-Z-Leto [24]

Answer:

a) $T(x)=300x+18x=318x$

b) $D(x)=318x-(0.1)(318)+10x\\=296.2x$

c) Yes.

Step-by-step explanation:

We have that:

$Tires=$300\\Tax=0.006\\Discount=0.1\\$

And we have $10 free of taxes.

Making x= number of tires to buy, then we have that the total cost of tires is:

$Total_{Tires}=300x$

So, what we pay for taxes is given by:

$Taxes=(300x)(0.06)=18x$

a) Then, according to the above, we can write down the total cost before the discount as:

$T(x)=300x+18x=318x$

b) And the total cost after discounts, is then given by:

$D(x)=318x-(0.1)(318)+10x\\=296.2x$

c) If the discount is added first, then less tax will be paid because the amount on which it is paid is lower. If the discount is added later, then the taxes will have been taxed on a higher amount, so it does matter whether they are added first or later.

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3 years ago

Polymer composite materials have gained popularity because they have high strength to weight ratios and are relatively easy and

Brums [2.3K]

Answer:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$

$p_v =P(t_9>10.349)=1.34x10^{-6}$

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=51.6$ represent the sample mean

$s=1.1$ represent the standard deviation for the sample

$n=10$ sample size

$\mu_o =48$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the mean score is higher than 48, the system of hypothesis would be:

Null hypothesis: $\mu \geq 48$

Alternative hypothesis: $\mu > 48$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{51.6-48}{\frac{1.1}{\sqrt{10}}}=10.349$

Calculate the P-value

First we find the degrees of freedom:

$df=n-1=10-1=9$

Since is a one-side upper test the p value would be:

$p_v =P(t_9>10.349)=1.34x10^{-6}$

Conclusion

If we compare the p value and the significance level given for example $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the the actual mean is significantly higher than 48 MPa at 5% of significance.

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