Name two ways you might see decimals used outside of school.

Consider the following sample of observations on coating thickness for low-viscosity paint.

Julli [10]

Answer:

a) $\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And for this case if we use this formula we got:

$\bar x = 1.3538$

b) Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

$Median = \frac{1.31+1.46}{2}= 1.385$

c) $P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.

d) $Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$

Step-by-step explanation:

Data given:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

Part a

We can calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And for this case if we use this formula we got:

$\bar x = 1.3538$

Part b

For this case in order to calculate the median we need to put the data on increasing way like this:

0.86 0.88 0.88 1.07 1.09 1.17 1.29 1.31 1.46 1.49 1.59 1.62 1.65 1.71 1.76 1.83

Since we have n =16 values for the sample the median can be calculated as the average between position 8th anf 9th and we got:

$Median = \frac{1.31+1.46}{2}= 1.385$

Part c

For this case we can assume that the mean is $\mu = 1.3538$

And we can calculate the population deviation with the following formula:

$\sigma = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{N}}$

And if we replace we got: $\sigma= 0.3105$

And assuming normal distribution we have this:

$X \sim N (\mu = 1.3538, \sigma= 0.3105)$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=1.3538 +1.28*0.3505=1.8024$

So the value of height that separates the bottom 90% of data from the top 10% is 1.8024.

Part d

The median is defined as :

$Median= \frac{x_{8} +x_{9}}{2}$

The variance for this estimator is given by:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} Var(X_{8} +X_{9})$

We can assume the obervations independent so then we have:

$Var(\frac{x_{8} +x_{9}}{2}) = \frac{1}{4} (2\sigma^2) = \frac{\sigma^2}{2}$

And replacing we got:

$Var(\frac{x_{8} +x_{9}}{2})= \frac{0.3105^2}{2}= 0.0482$

And the standard error would be given by:

$Sd(\frac{x_{8} +x_{9}}{2})= \sqrt{0.0482}=0.2196$

6 0

2 years ago

Jonah has a recipe that uses 1 1/2 cups of brown sugar and 2 1/3 cups of flour to make 24 muffins has a total of 7 cups of flour

Mekhanik [1.2K]

Answer:

aaaa

Step-by-step explanation:a

6 0

2 years ago

Only 35% of the drivers in a particular city wear seat belts. Suppose that 20 drivers are stopped at random what is the probabil

lesya692 [45]

Here we have a situation where the probability of a driver wearing seat belts is known and remains constant throughout the experiment of stopping 20 drivers.

The drivers stopped are assumed to be random and independent.

These conditions are suitable for modelling using he binomial distribution, where

where n=number of drivers stopped (sample size = 20)

x=number of drivers wearing seatbelts (4)

p=probability that a driver wears seatbelts (0.35), and

C(n,x)=binomial coefficient of x objects chosen from n = n!/(x!(n-x)!)

So the probability of finding 4 drivers wearing seatbelts out of a sample of 20

P(4;20;0.35)

=C(20,4)*(0.35)^4*(0.65)^16

= 4845*0.0150061*0.0010153

= 0.07382

6 0

2 years ago

The Jacksons and the Simpsons were competing in the final leg of the Amazing Race, which was 240240240 kilometers long. In their

meriva

Answer:

240/(v+40) + 1 > 240/v

Step-by-step explanation:

Both parties traveled a distance of 240 km.

Simpsons took off at average speed of 40 km/h & 1 hour later: 240/(v+40) + 1

Jacksons took off at average speed of v km/h: 240/v

6 0

2 years ago

In a school of 330 students, 85 of them are in the drama club, 200 of them are in a sports team and 60 students do drama and spo

liq [111]

Let's denote students in D = in the drama club , S = in a sports team 
P(D) = 85/330 
P(S) = 200/330 

P(D and S) = 60/300 

P(D or S) = P(D) + P(S) - P(D and S) = 85/330 + 200/330 - 60/330 = 15/22 

P(neither D nor S) = 1- 15/22 = 7/22

7 0

2 years ago