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Snowcat [4.5K]
3 years ago
10

WORKOUT

Mathematics
1 answer:
SpyIntel [72]3 years ago
4 0

Answer:

= 6 typists

Step-by-step explanation:

Well, 2 can type 2 in 2 mins,

So, 1 can type 2 in 4 mins.

So, 1 can type 18 in 36 mins.

So, 2 can type 18 in 18 mins.

So, 6 can type 18 in 6 mins.

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A random sample of 400 voters in a certain city are asked if they favor an additional 4% gasoline tax to provide badly needed re
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Answer:

A) α = 0.04136

B) β = 0.00256

Step-by-step explanation:

We are given;

Sample size; n = 400

Proportion; p = 60% = 0.6

Formula for mean is;

μ = np

μ = 400 × 0.6 = 240

Standard deviation is given by;

σ = √npq

Where q = 1 - p = 1 - 0.6 = 0.4

σ = √(400 × 0.6 × 0.4)

σ = √96

σ = 9.8

A) our null hypothesis is at p = 0.6

Probability of making a type I error means we reject the null hypothesis when it is true.

This can be expressed in reference to the question as;

α = P(x < 220) + P(x > 260) all at p = 0.6

Now,

P(x < 220) = z = (x¯ - μ)/σ = (220 - 240)/9.8 = -2.04

Also;

P(x > 260) = z = (260 - 240)/9.8 = 2.04

Now, from z-distribution table probability of a z-score of -2.04 is 0.02068.

Also, probability of z-score of 2.04 is (1 - P(z < 2.04) = 1 - 0.97932 = 0.02068

Thus;

α = 0.02068 + 0.02068

α = 0.04136

B) Type II error occurs when we fail to reject the null hypothesis even though it's false.

In this case our alternative hypothesis is at p = 48% = 0.48

Thus;

μ = np

μ = 400 × 0.48 = 192

Standard deviation is given by;

σ = √npq

Where q = 1 - p = 1 - 0.48 = 0.52

σ = √(400 × 0.48 × 0.52)

σ = √99.84

σ = 9.992

Type II error would be given by;

β = [((x1¯ - μ)/σ) < z > ((x2¯ - μ)/σ)]

β = [((220 - 192)/9.992) < z > ((260 - 192)/9.992)]

β = (2.8 < z > 6.81)

Rearranging this gives us;

β = P(z < 6.81) - P(z < 2.8)

From z-distribution tables, we have;

β = 1 - 0.99744

β = 0.00256

3 0
3 years ago
For​ $3.98 you can get a​ salad, main​ course, and dessert at the cafeteria. If you have a choice of 2 different​ salads, 2 diff
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I think totoly 2 because u need all three of them.
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At one point the average price of regular unleaded gasoline was ​$3.39 per gallon. Assume that the standard deviation price per
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This question was not written completely

Complete Question

At one point the average price of regular unleaded gasoline was ​$3.39 per gallon. Assume that the standard deviation price per gallon is ​$0.07 per gallon and use​ Chebyshev's inequality to answer the following.

​(a) What percentage of gasoline stations had prices within 3 standard deviations of the​ mean?

​(b) What percentage of gasoline stations had prices within 2.5 standard deviations of the​ mean? What are the gasoline prices that are within 2.5 standard deviations of the​ mean?

​(c) What is the minimum percentage of gasoline stations that had prices between ​$3.11 and ​$3.67​?

Answer:

a) 88.89% lies with 3 standard deviations of the mean

b) i) 84% lies within 2.5 standard deviations of the mean

ii) the gasoline prices that are within 2.5 standard deviations of the​ mean is $3.215 and $3.565

c) 93.75%

Step-by-step explanation:

Chebyshev's theorem is shown below.

1) Chebyshev's theorem states for any k > 1, at least 1-1/k² of the data lies within k standard deviations of the mean.

As stated, the value of k must be greater than 1.

2) At least 75% or 3/4 of the data for a set of numbers lies within 2 standard deviations of the mean. The number could be greater.μ - 2σ and μ + 2σ.

3) At least 88.89% or 8/9 of a data set lies within 3 standard deviations of the mean.μ - 3σ and μ + 3σ.

4) At least 93.75% of a data set lies within 4 standard deviations of the mean.μ - 4σ and μ + 4σ.

​

(a) What percentage of gasoline stations had prices within 3 standard deviations of the​ mean?

We solve using the first rule of the theorem

1) Chebyshev's theorem states for any k > 1, at least 1-1/k² of the data lies within k standard deviations of the mean.

As stated, the value of k must be greater than 1.

Hence, k = 3

1 - 1/k²

= 1 - 1/3²

= 1 - 1/9

= 9 - 1/ 9

= 8/9

Therefore, the percentage of gasoline stations had prices within 3 standard deviations of the​ mean is 88.89%

​(b) What percentage of gasoline stations had prices within 2.5 standard deviations of the​ mean?

We solve using the first rule of the theorem

1) Chebyshev's theorem states for any k > 1, at least 1-1/k² of the data lies within k standard deviations of the mean.

As stated, the value of k must be greater than 1.

Hence, k = 3

1 - 1/k²

= 1 - 1/2.5²

= 1 - 1/6.25

= 6.25 - 1/ 6.25

= 5.25/6.25

We convert to percentage

= 5.25/6.25 × 100%

= 0.84 × 100%

= 84 %

Therefore, the percentage of gasoline stations had prices within 2.5 standard deviations of the​ mean is 84%

What are the gasoline prices that are within 2.5 standard deviations of the​ mean?

We have from the question, the mean =$3.39

Standard deviation = 0.07

μ - 2.5σ

$3.39 - 2.5 × 0.07

= $3.215

μ + 2.5σ

$3.39 + 2.5 × 0.07

= $3.565

Therefore, the gasoline prices that are within 2.5 standard deviations of the​ mean is $3.215 and $3.565

​(c) What is the minimum percentage of gasoline stations that had prices between ​$3.11 and ​$3.67​?

the mean =$3.39

Standard deviation = 0.07

Applying the 2nd rule

2) At least 75% or 3/4 of the data for a set of numbers lies within 2 standard deviations of the mean. The number could be greater.μ - 2σ and μ + 2σ.

the mean =$3.39

Standard deviation = 0.07

μ - 2σ and μ + 2σ.

$3.39 - 2 × 0.07 = $3.25

$3.39 + 2× 0.07 = $3.53

Applying the third rule

3) At least 88.89% or 8/9 of a data set lies within 3 standard deviations of the mean.μ - 3σ and μ + 3σ.

$3.39 - 3 × 0.07 = $3.18

$3.39 + 3 × 0.07 = $3.6

Applying the 4th rule

4) At least 93.75% of a data set lies within 4 standard deviations of the mean.μ - 4σ and μ + 4σ.

$3.39 - 4 × 0.07 = $3.11

$3.39 + 4 × 0.07 = $3.67

Therefore, from the above calculation we can see that the minimum percentage of gasoline stations that had prices between ​$3.11 and ​$3.67​ corresponds to at least 93.75% of a data set because it lies within 4 standard deviations of the mean.

4 0
4 years ago
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