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"Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for metal sheets of a particular type, its mean value and standard deviation are 75 GPa and 2.1 GPa, respectively. Suppose the distribution is normal. (Round your answers to four decimal places.)(a) Calculate P(74 ? X ? 76) when n = 25.(b)How likely is it that the sample mean diameter exceeds 76 when n = 49?"

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Answer 1

Answer:

Answer:

a) $P(74 <\bar X <76)=P(-2.38< Z< 2.38) = P(Z<2.38)-P(Z<-2.38)=0.9913-0.0087=0.9827$

b) $P(\bar X >76)=P(Z>3.33)=1-P(Z<3.33)=1-0.9996=0.0004$

So on this case is very unlikely that the sample mean would be higher exceeds 76 when n =49

Step-by-step explanation:

1) 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".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

2) Part a

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

$X \sim N(75,2.1)$

Where $\mu=75$ and $\sigma=2.1$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=75, (\sigma)/(√(n))=(2.1)/(√(25)))$

And we want to find this probability:

$P(74 <\bar X <76)=P((74-75)/((2.1)/(√(25)))<Z<(76-75)/((2.1)/(√(25))))$

$P(-2.38< Z< 2.38) = P(Z<2.38)-P(Z<-2.38)=0.9913-0.0087=0.9827$

3) Part b

For this case the new distribution is given by:

$\bar X \sim N(\mu=75, (\sigma)/(√(n))=(2.1)/(√(49)))$

$P(\bar X >76)=P(Z>(76-75)/((2.1)/(√(49)))=3.33)$

$P(Z>3.33)=1-P(Z<3.33)=1-0.9996=0.0004$

So on this case is very unlikely that the sample mean would be higher exceeds 76 when n =49

Answer 2

Answer:

Final answer:

In this problem, we calculated the probabilities for a normal distribution with mean 75 and standard deviation 2.1. For part (a), the probability was found to be 0.9179 that the modulus lies between 74 and 76 for a sample size of 25. In part (b), the likelihood was found to be 0.0004 that the sample mean diameter of 49 pieces exceeds 76.

Explanation:

Given the problem, we are dealing with a normal distribution with a mean of 75 GPA and a standard deviation of 2.1 GPA.

(a) We are asked to find the probability that the modulus is between 74 and 76 for a sample size of 25. First, we calculate the standard deviation of the sample mean as sigma/sqrt(n), which is 2.1/sqrt(25)=0.42. Then, we find the Z-scores for 74 and 76 as (X - mean)/standard deviation, yielding -2.3809 and 2.3809. We lookup these Z-scores in a Z-table, find their corresponding probabilities, and subtract to get 0.9179.

(b) Next, we want to find how likely the sample mean diameter exceeds 76 GPA for n = 49. We find the new standard deviation as 2.1/sqrt(49)=0.3. The Z-score for 76 is (76 - 75)/0.3 =3.3333. We subtract the corresponding probability (0.9996) in the Z-table from 1 to find it's incredibly likely at 0.0004 that the sample mean diameter exceeds 76.

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D) 3

carollnx talk me

............ that is the correct options

Answer:

D) 3

be confident to answer any questions

which of the following is a Quadrilateral which must contain diagonals that are always congruent and must always have 90° angles?

Answers

Answer: B

Step-by-step explanation:

The commute times (in minutes) of 30 employees are listed below. (a) Find Qi, Q3, and the interquartile range (IQR).
(b) Find the fences and determine if there are any outliers in the sample.
20 4 45 48 52 55 56 60 63 65 67 68 6 70 74 75 77 78 79 80 81 82 8 85 87 88 90 92 95 99

Answers

Answer:

Q1 = 54.25 ; Q3 = 82.75 ; IQR = 28.5 ; Outliers = 4, 6, 8

Step-by-step explanation:

Quartiles divide data into 4 equal parts. Q1 has 25% data below it & 75% above it, Q3 has 75% data below it & 25% data above it.

Quartile Calculation for 30 number of observations

Ascending order : 4, 6, 8, 20, 45, 48, 52, 55, 56, 60, 63, 65, 67, 68, 70, 74, 75, 77, 78, 79, 80, 81, 82, 85, 87, 88, 90, 92, 95, 99
Q1 = (N + 1) / 4th observation = [ (30 + 1)/4 ] th obs. = (31 /4) th = 7.75th = 7th item + 0.75 (7th - 8th item) = 52 + 0.75 (55-52) = 52 + 0.75(3) = 52 + 2.25 = 54.25
Q3 = 3 [(N+1) / 4] th observation = 7.75 x 3 = 23.25th item = 23rd item + 0.25 (24th - 23rd item) = 82 + 0.25 (85-82) = 82 + 0.25 (3) = 82 + 0.75 = 82.75
Interquartile range (IQR) = Q3 - Q1 = 82.75 - 54.25 = 28.5

Outliers are the observations less than Q1 - 1.5 (IQR) , more than Q3 + 1.5 (IQR)

Q1 - 1.5 (IQR) = 54.25 - 1.5 (28.5) = 54.25 - 42.75 = 11.5 .So, numbers less than 11.5, ie 4, 6, 8 are outliers
Q3 + 1.5 (IQR) = 82.75 + 1.5 (28.5) = 82.75 + 42.75 = 125.5 .There are no numbers > 125.5

Final answer:

To find Qi, Q3, and the interquartile range (IQR), first, the data needs to be sorted in ascending order. Then, find Q1 as the median of the lower half and Q3 as the median of the upper half. The IQR is the difference between Q3 and Q1. To determine if there are any outliers, calculate the fences using the formulas: Lower fence = Q1 - (1.5 * IQR) and Upper fence = Q3 + (1.5 * IQR). If any commute times fall outside the fences, they are considered outliers.

Explanation:

To find the quartiles, the data needs to be sorted in ascending order. The sorted list is:

4, 6, 8, 20, 45, 48, 52, 55, 56, 60, 63, 65, 67, 68, 70, 74, 75, 77, 78, 79, 80, 81, 82, 85, 87, 88, 90, 92, 95, 99

Q1 is the median of the lower half (ignoring the median if the number of data points is odd), so Q1 = 56.

Q3 is the median of the upper half, so Q3 = 81.

The interquartile range (IQR) is the difference between Q3 and Q1, so IQR = Q3 - Q1 = 81 - 56 = 25.

To find the fences, use the formulas:

Lower fence: Q1 - (1.5 * IQR)

Upper fence: Q3 + (1.5 * IQR)

Lower fence = 56 - (1.5 * 25) = 56 - 37.5 = 18.5

Upper fence = 81 + (1.5 * 25) = 81 + 37.5 = 118.5

There are no commute times that fall outside the fences, so there are no outliers in the sample.

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