The top-selling red and voss tire is rated 7 miles, which means nothing. in fact, the distance the tires can run until wear-out is a normally distributed random variable with a mean of 79000 miles and a standard deviation of 5000 miles. a. what is the probability that the tire wears out before 7 miles? probability = 0.0539 b. what is the probability that a tire lasts more than 83000 miles? probability =

Step-by-step explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.

Margin of error = z × σ/√n

Where

σ = population standard Deviation

n = number of samples

From the information given

1) x = 17.5

σ = 5

n = 50

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.9 = 0.1

α/2 = 0.1/2 = 0.05

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.05 = 0.95

The z score corresponding to the area on the z table is 1.645. Thus, confidence level of 90% is 1.645

Margin of error = 1.645 × 5/√50 = 1.16

Confidence interval = 17.5 ± 1.16

3.92% probability that, for an adult after a 12-hour fast, x is less than 53

Step-by-step explanation:

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

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

In this problem, we have that:

[tex]\mu = 90, \sigma = 21[/tex]

What is the probability that, for an adult after a 12-hour fast, x is less than 53?

This is the pvalue of Z when X = 53.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{53 - 90}{21}[/tex]

[tex]Z = -1.76[/tex]

[tex]Z = -1.76[/tex] has a pvalue of 0.0392

3.92% probability that, for an adult after a 12-hour fast, x is less than 53

Uniu-nos há há um uma i

16%

Step-by-step explanation:

The indicated probability is actually the area under the standard normal curve to the left of the mean. I used the function normalcdf( on my TI-83 Plus calculator to find this quantity:

normalcdf(-1000,85,90,5) = 0.1587.

Note #1: This quantity (area / probability) is the area to the left of 85.

Note #2: by the Empirical Rule, 68% of data lies within 1 s. d. of the mean, so the area between the mean (90) and the score (85) is half of 68%, or 34%. Subtracting this from 50% (the area to the left of the mean), we get 16%, which is roughly equivalent to the 0.1587 we got earlier.

a) [tex]\sigma_{\bar X}= \frac{1.25}{\sqrt{16}}= 0.3125[/tex]

b) [tex]25-1.64\frac{1.25}{\sqrt{16}}=24.4875[/tex]

[tex]25+1.64\frac{1.25}{\sqrt{16}}=25.5125[/tex]

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(25,1.25)[/tex]

Where [tex]\mu=25[/tex] and [tex]\sigma=1.25[/tex]

Part a

We know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

The standard error is given by:

[tex]\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}[/tex]

And replacing we got:

[tex]\sigma_{\bar X}= \frac{1.25}{\sqrt{16}}= 0.3125[/tex]

Part b

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

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=1.64[/tex]

Now we have everything in order to replace into formula (1):

[tex]25-1.64\frac{1.25}{\sqrt{16}}=24.4875[/tex]

[tex]25+1.64\frac{1.25}{\sqrt{16}}=25.5125[/tex]

Step-by-step explanation:

The distance the tyres can run until wear-out is a normally distributed, the formula for normal distribution is expressed as

z = (x - u)/s

Where

x = the distance the tyres can run until wear-out

u = mean distance

s = standard deviation

From the information given,

u = 7900 miles

s = 5000 miles

A) We want to find the probability probability that the tyre wears out before 70000 miles. It is expressed as

P(x lesser than 70000)

For x = 70000,

z = (70000 - 79000)/5000 = - 1.8

Looking at the normal distribution table, the corresponding z score is 0.03593

B)We want to find the probability probability that the tyre lasts more than 83000 miles. It is expressed as

P(x greater than 83000) = 1 - P(x lesser than or equal to 83000)

For x = 83000,

z = (83000 - 79000)/5000 = 0.8

Looking at the normal distribution table, the corresponding z score is 0.78814

P(x greater than 83000)

= 1 - 0.78814 = 0.21

[tex](9.27025,\ 11.12975)[/tex]

Step-by-step explanation:

Given : Sample size : n= 9

Degree of freedom = df =n-1 =8

Sample mean : [tex]\overline{x}=10.2[/tex]

sample standard deviation : [tex]s= 1.5[/tex]

Significance level ; [tex]\alpha= 1-0.90=0.10[/tex]

Since population standard deviation is not given , so we use t- test.

Using t-distribution table , we have

Critical value = [tex]t_{\alpha/2, df}=t_{0.05 , 8}=1.8595[/tex]

Confidence interval for the population mean :

[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]

90% confidence interval for the mean value will be :

[tex]10.2\pm (1.8595)\dfrac{1.5}{\sqrt{9}}[/tex]

[tex]10.2\pm (1.8595)\dfrac{1.5}{3}[/tex]

[tex]10.2\pm (1.8595)(0.5)[/tex]

[tex]10.2\pm (0.92975)[/tex]

[tex](10.2-0.92975,\ 10.2+0.92975)[/tex]

[tex](9.27025,\ 11.12975)[/tex]

Hence, the 90% confidence interval for the mean value= [tex](9.27025,\ 11.12975)[/tex]