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Lena [83]
3 years ago
14

Determine the point estimate of the population proportion, the margin of error

Mathematics
1 answer:
ahrayia [7]3 years ago
8 0

Answer:

\hat p=0.645\\\\ME=0.233\\\\x=645 \ individuals

Step-by-step explanation:

-Given the boundaries as 0.412 and 0.878

-\hat p is the point estimate for the population proportion and is calculated as follows:

\hat p=\frac{Upper \ Bound+Lower \ Bound}{2}\\\\=\frac{0.878+0.412}{2}\\\\=0.645\\\\

#The margin of error, ME can be calculated for the confidence intervals using the formula:

ME=\frac{Upper \ Bound-Lower \ Bound}{2}\\\\=\frac{0.878-0.412}{2}\\\\=0.233

#The number of individuals in the sample is the product of the point estimate and population size:

\hat p=\frac{x}{n}\\\\x=\hat pn\\\\=0.645\times 1000\\\\=645

Hence, there are 645 individuals in the sample.

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Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
3 years ago
-9 2/3(-2 1/4a+b+8 1/4)<br><br> Solve please with work
harkovskaia [24]

Answer:

4/x-1 -5/x+4/4/x-14/x-14/x-1 -5/x+2=3/x -5/x+2=3/x -5/x+2=3/xx-1 -5/x+2=3/x2=3/x

3 0
3 years ago
The distance around the middle of a sphere (the circumference) equals 37.699 in. what is the surface area of the sphere?
podryga [215]

To solve this problem you must apply the proccedure shown below:

1. Clear the radius from the formula for the circumference of a circle and substitute values, as following:

C=2\pi r\\ r=C/2\pi \\ r=37.699in/2\pi \\ r=5.99in

2. Substitute the radius calculated into the formula for calculate the surface area of the sphere:

SA=4\pi r^{2} \\ SA=4\pi(5.99in^{2})\\ SA=450.88in^{2}

The answer is: 450.88in^{2}

5 0
3 years ago
Use slopes and y-intercepts to determine if the lines 10x+3y=−3 and 5x−4y=−3 are parallel.
kvasek [131]

Answer:

They are not parallel

Step-by-step explanation:

original equation

10x + 3y = -3

subtract 10x

3y = -10x - 3

divide by 3

y = -10/3x - 1

original equation

5x - 4y = -3

subtract 5x

-4y = -5x-3

divide by -4

y = 5/4x + 3/4

the slopes are not equal to each other (5/4x and -10/3x) so they are not parallel

4 0
1 year ago
PLZ ANSWER ASAP!!!
ruslelena [56]
One because if you look at the graph you go to 3 and there is only one.
4 0
3 years ago
Read 2 more answers
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