When you divide any number by a fraction less than one, how does the original number change?

Answer:

[tex]z=1.28<\frac{a-490}{10}[/tex]

And if we solve for a we got

[tex]a=490 +1.28*10=502.8[/tex]

So the value of height that separates the bottom 90% of data from the top 10% is 502.8.  

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 weeknly amount of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(490,10)[/tex]  

Where [tex]\mu=490[/tex] and [tex]\sigma=10[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X>a)=0.1[/tex]   (a)

[tex]P(X<a)=0.9[/tex]   (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:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.9[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.9[/tex]

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

[tex]z=1.28<\frac{a-490}{10}[/tex]

And if we solve for a we got

[tex]a=490 +1.28*10=502.8[/tex]

So the value of height that separates the bottom 90% of data from the top 10% is 502.8.  

Final answer:

The company should budget approximately $502.80 for weekly repairs and maintenance to ensure that the probability of exceeding this amount is only 0.1.

Explanation:

We want to find how much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1.

Since the weekly amount of money spent is normally distributed with a mean of $490 and a standard deviation of $10, we can find the amount by looking up the z-score that corresponds to the 90th percentile

(since 100% - 10% = 90%) in a standard normal distribution table or using a calculator.

Let the z-score for the 90th percentile be denoted as z.

Looking up the standard normal distribution table or using a calculator, we find that z ≈ 1.28 for 0.9 cumulative probability.

We then use the z-score formula:

z = (X - mean) / standard deviation

Plugging in our z-score and the parameters, we can solve for X:

1.28 = (X - 490) / 10

X - 490 = 12.8

X = $502.80

Therefore, the company should budget approximately $502.80 for weekly repairs and maintenance to ensure that the probability of exceeding this amount is only 0.1.

Answer:

5(8-x)

Step-by-step explanation:

Fact using the GCF (Greatest Common Factor) of 40 and 5

Factors of 5:

1,5

Factors of 40:

1,2,4,5,8,10,20,40

The greatest factor they both have is 5

Factor by distributing out a 5

40-5x

5(8-x)

The GCF (Greatest Common Factor) of 40 and 5 is 5!

Divide both numbers by 5.

40 / 5 = 8

-5x / 5 = -x

Rewrite the expression using the distributive property.

5(8 - x)

Best of Luck!

Answer:

1/4; 25%

Step-by-step explanation:

Both events happen with probability 1/2: there are 3 even numbers and 3 odd numbers in a die.

Since the two events are also independent (the outcome of the first die doesn't affect the outcome of the second), we have to multiply those probability.

So, you roll an odd number on the first die and an even number on the second die with probability

[tex]\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}[/tex]

Answer:

[tex]\frac{33}{36}[/tex]

Step-by-step explanation:

Combinations greater than a 10

5 - 5

5 - 6

6 - 5

There are total 36 combinations (6 * 6).

3 of these combinations are higher than 10.

So 36 - 3 combinations are less than 10.

Answer: 35.5cm^2

Step-by-step explanation:

I believe we're talking about a rectangular. So it would be length * width or 7.5cm * 5cm = 35.5cm^2

Answer:

1. Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex]  70 mph

  Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 70 mph

2. Value of test statistics is 2.652.

Step-by-step explanation:

We are given that Kansas Highway Patrol recorded the speed of 450 interstate vehicles and found the mean speed of 70.2 mph with standard deviation 1.6 mph.

We have to conduct a hypothesis test at significance level 0.01 to decide whether or not to reduce the money sent to Kansas.

Let [tex]\mu[/tex] = true mean speed of all vehicles on the Kansas interstate highway system.

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex]  70 mph   {means that the federal government will not reduce the amount of funding it provides as the speed limit is less than or equal to 70 mph}

Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] > 70 mph   {means that the federal government will reduce the amount of funding it provides as the speed limit exceed 70 mph}

The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

                        T.S.  = [tex]\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }[/tex]  ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean speed limit of 450 interstate vehicles = 70.2 mph

             s = sample standard deviation = 1.6 mph

             n = sample of vehicles = 450

So, test statistics  =   [tex]\frac{70.2-70}{\frac{1.6}{\sqrt{450} } }[/tex]  ~ [tex]t_4_4_9[/tex]

                               =  2.652

Hence, the value of test statistics is 2.652.

Answer:

y=6x+7

Step-by-step explanation:

−6x+y=7

Step 1: Add 6x to both sides

y=6x+7

Answer:

y=6x+7

Step-by-step explanation:

whenever you solve for a letter you have to get it by itself. That means it has to be alone on the opposite side of the equation

move the -6x to the other side of the equation

that -6 changes to a  positive 6 because, your moving it to the other side of the equation

your left with y=6x+7

pls mark me brainliest

Answer:

17.38m

Step-by-step explanation:

Answer:

17.38m

Step-by-step explanation:

Answer:

The answers are for option (a) 0.2070  (b)0.3798  (c) 0.3938

Step-by-step explanation:

Given:

Here Section 1 students = 20

Section 2 students = 30

Here there are 15 graded exam papers.

(a )Here Pr(10 are from second section) = ²⁰C₅ * ³⁰C₁₀/⁵⁰C₁₅= 0.2070

(b) Here if x is the number of students copies of section 2 out of 15 exam papers.

 here the distribution is hyper-geometric one, where N = 50, K = 30 ; n = 15

Then,

Pr( x ≥ 10 ; 15; 30 ; 50) = 0.3798

(c) Here we have to find that at least 10 are from the same section that means if x ≥ 10 (at least 10 from section B) or x ≤ 5 (at least 10 from section 1)

so,

Pr(at least 10 of these are from the same section) = Pr(x ≤ 5 or x ≥ 10 ; 15 ; 30 ; 50) = Pr(x ≤ 5 ; 15 ; 30 ; 50) + Pr(x ≥ 10 ; 15 ; 30 ; 50) = 0.0140 + 0.3798 = 0.3938

Note : Here the given distribution is Hyper-geometric distribution

where f(x) = kCₓ)(N-K)C(n-x)/ NCK in that way all these above values can be calculated.

Answer:

The theoretical probability of landing on 2 heads, when 10 coins are tossed is 0.0439 or 4.39%.

Step-by-step explanation:

Number of coins that fell on the floor = 10

Number of coins that landed on heads = 2

We have to find the theoretical probability of getting 2 coins landing of heads when 10 coins are tossed.

Notice that there are only 2 possible outcomes: Either that coin will land on head or it won't. Landing of each coin is independent of the others coins. Probability of each coin landing on head is constant i.e. 0.5 or 1/2. Number of trials, i.e. the number of times the experiment will be done is fixed, which is 10.

All the 4 conditions for an experiment to be considered a Binomial Experiment are satisfied. So we will use Binomial Probability to solve this problem.

Probability of success = Probability of coin landing on head = 0.5

Number of trials = n = 10

Number of success = r = 2

The formula for Binomial Probability is:

[tex]P(X = x) =^{n}C_{r}(p)^{r}(1-p)^{n-r}[/tex]

Using the values, we get:

[tex]P(X=2)=^{10}C_{2}(0.5)^2(0.5)^8=0.0439[/tex]

Thus, the theoretical probability of landing on 2 heads, when 10 coins are tossed is 0.0439 or 4.39%.

Answer: a) 1.029, b) (-5.318, -1.282).

Step-by-step explanation:

Since we have given that

[tex]n_1=13\\\\n_2=17\\\\\bar{x_1}=1.9\\\\\bar{x_2}=5.2[/tex]

and

[tex]s_1=2.1\\\\s_2=3.5[/tex]

So, the standard error for comparing the means :

[tex]SE=\sqrt{\dfrac{s^2_1}{n_1}+\dfrac{s^2_2}{n_2}}\\\\SE=\sqrt{\dfrac{2.1^2}{13}+\dfrac{3.5^2}{17}}\\\\SE=\sqrt{1.0598}\\\\SE=1.029[/tex]

At 95% confidence interval, z = 1.96

So, Confidence interval would be

[tex]\bar{x_1}-\bar{x_2}\pm z\times SE\\\\=(1.9-5.2)\pm 1.96\times 1.0294\\\\=-3.3\pm 2.017624\\\\=(-3.3-2.018,-3.3+2.018)\\\\=(-5.318,-1.282)[/tex]

Hence, a) 1.029, b) (-5.318, -1.282).

Answer:

so the common multiple of 4 and 6 is 2  i hope this helps!   :)

Step-by-step explanation:

well the common multiples of 4 and 6 is 2

2 * 2 = 4

2 * 3 = 6

so there has to be a certain number that goes into both of the numbers you are working with

Answer:

Common multiples of 4 and 6 are 12, 24, 36, 48, 60, 72, 84, 96...

Final answer:

To test whether the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer, we can use a one-sample proportion test. Using a significance level of 0.05, we find that the test statistic is -2.09. Comparing this to the critical value from the standard normal distribution (-1.645), we reject the null hypothesis and conclude that there is evidence to suggest that the percentage of correctly guessed penalty shots is less than 50%.

Explanation:

To test whether there is evidence that the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer, we can use a one-sample proportion test. We will assume that the null hypothesis is true and that the goalkeeper's correct guesses are no better than random chance. The alternative hypothesis would be that the goalkeeper's correct guesses are significantly less than 50%. Using a significance level of 0.05, we can calculate the test statistic and compare it to the critical value from the standard normal distribution.

Null hypothesis (H0): The percentage of correctly guessed penalty shots is 50%.

Alternative hypothesis (Ha): The percentage of correctly guessed penalty shots is less than 50%.

Test statistic: We can use the z-test statistic since the sample size is large enough. The formula for the z-test statistic is z = (p - P0) / SE, where p is the sample proportion, P0 is the hypothesized proportion, and SE is the standard error. In this case, since the standard error is given as 0.043, we can plug in the values to calculate the test statistic.

Calculate the z-test statistic: z = (0.41 - 0.5) / 0.043 = -2.09

Find the critical value: Since our alternative hypothesis is that the percentage is less than 50%, we will use a one-tailed test. With a significance level of 0.05, the critical value from the standard normal distribution is -1.645.

Compare the test statistic to the critical value: Since the test statistic (-2.09) is less than the critical value (-1.645), we can reject the null hypothesis. There is evidence to suggest that the percentage of correctly guessed penalty shots is less than 50% in World Cup Soccer.

Answer:EX = RT ln [X]o

.........zF.....[X]i

EX = (1.987 cal/deg.mol)(293 deg) ln [X]o

.........z(23,062 cal/volt.mol)................[X]i

OR

EX = (8.315 joules/deg.mol)(293 deg) ln [X]o

.........z(96,485 joules/volt.mol)................[X]i

EK+ = 0.025 ln(12/400) = -0.088 V = -88 mV

ENa+ = 0.025 ln(450/55) = 0.053 V = 53 mV

ECa+2 = 0.0126 ln(10/0.0001) = 0.145 V = 145 mV

ECl- = -0.025 ln(550/56) = -0.058 V = -58 mV

Step-by-step explanation:

The Goldman equation is used to calculate resting membrane potential (Vm) considering the relative permeabilities of different ions. Given the permeabilities, the resting Vm can be estimated. If the K+ permeability increases, Vm will move closer to the Potassium equilibrium potential (EK).

The resting membrane potential (Vm) can be calculated using the Goldman equation, which considers the relative permeabilities and concentrations of different ions. The equation is: Vm = 61.5 log ((PK[K+]out + PNa[Na+]out + PCl[Cl-]in) / (PK[K+]in + PNa[Na+]in + PCl[Cl-]out)), where Px indicates the relative permeability of each ion & the square brackets contain the ion concentrations inside (in) or outside (out) the cell.

Given the permeabilities PK : PNa : PCl = 1.0 : 0.04 : 0.45, assuming concentrations inside and outside the cell in a balanced condition with no net movement of any ion, you might estimate the resting Vm.  

If the K+ permeability increases, Vm would reportedly move closer towards the Potassium equilibrium potential (EK). This is because the membrane is becoming more permeable to K+ and less responsive to the influences of other ions.

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Answer:

(a) The outlier in the data is $810.

(b) The distribution of money spent on textbooks for the 34 students is right skewed.

Step-by-step explanation:

The data provided for the amount of money 34 college students spent on books is:

S = {120, 130, 130, 140, 150, 150, 160, 170, 180, 210, 220, 230, 240, 250, 260, 280, 280, 290, 290, 290, 310, 320, 320, 370, 380, 390, 410, 440, 450, 470, 510, 530, 620, 810}

(a)

An outlier of a data set is a value that is very different from the other values of a data set. It is either too large or too small.

The most common way to determine whether a data set consists of any outliers of not is,

Here

Q₁ = first quartile

Q₃ = third quartile

IQR = Inter-quartile range = Q₃ - Q₁.

The first quartile is the value that is more than 25% of the data values. The first quartile is the median of the first half of the data.

Compute the value of first quartile as follows:

First half of data: {120, 130, 130, 140, 150, 150, 160, 170, 180, 210, 220, 230, 240, 250, 260, 280, 280}

There are 17 values.

The median of an odd data set is the middle value.

The middle value is: 180

The first quartile is Q₁ = 180.

The third quartile is the value that is more than 75% of the data values.

Compute the value of first quartile as follows:

Second half of data: {290, 290, 290, 310, 320, 320, 370, 380, 390, 410, 440, 450, 470, 510, 530, 620, 810}

There are 17 values.

The median of an odd data set is the middle value.

The middle value is: 390

The third quartile is Q₃ = 390.

Compute the inter-quartile range as follows:

IQR = Q₃ - Q₁

      = 390 - 180

      = 210

Compute the value of [Q₁ - 1.5 IQR] as follows:

[tex]Q_{1}-1.5\ IQR =180-(1.5\times 210)=-135[/tex]

Compute the value of [Q₃ + 1.5 IQR] as follows:

[tex]Q_{3}+1.5\ IQR =390+(1.5\times 210)=705[/tex]

There are no values that are less than [Q₁ - 1.5 IQR]. But there is one value that is more than [Q₃ + 1.5 IQR].

X = 810 > [Q₃ + 1.5 IQR] = 705

Thus, the outlier in the data is $810.

(b)

A distribution is known as to be skewed to the right, or positively skewed, when maximum of the data are collected on the left of the distribution.

In the stem plot above, it is shown that maximum of the data values are collected on the left of the chart. This implies that the distribution is positively skewed.

Thus, the distribution of money spent on textbooks for the 34 students is right skewed.

Outliers are data elements that are relatively far from other data elements

(a) The outlier

From the plot, the stems are given as:

Stems: 1 2 3 4 5 6 7 8

However, stem 7 does not have any leaf

Using the above highlight, the next entry after 620 is 810

810 is relatively far from the other datasets.

Hence, 810 is an outlier

(b) The distribution

The data on the stem-plot is more concentrated at the top, and it reduces as the stem increases.

When there are more data elements at the top or left, then the distribution is right skewed.

Hence, the distribution is right skewed.

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Answer:

Option D) 0.65    

Step-by-step explanation:

We are given the following in the question:

Percentage of executives who read Time magazine = 35%

[tex]P(M) = 0.35[/tex]

Percentage of executives who read U.S. News & World Report = 40%

[tex]P(N) = 0.4[/tex]

Percentage of executives who read both Time magazine and U.S. News & World Report = 10%

[tex]P(M\cap N) = 0.1[/tex]

We have to find the probability that a particular top executive reads either Time or U.S. News & World Report regularly.

Thus, we have to evaluate,

[tex]P(M\cup N) = P(M) + P(N) -P(M\cap N)[/tex]

Putting values, we get,

[tex]P(M\cup N) = 0.35 + 0.4 - 0.1=0.65[/tex]

0.65 is the probability that a particular top executive reads either Time or U.S. News & World Report regularly.

Thus, the correct answer is

Option D) 0.65

Without the provided number line, we cannot determine the exact relationship between f and g. Inequalities f < g or f > g represent f being less than or greater than g, respectively. To write the correct inequality, one must refer to the positions of f and g on the number line.

Since the number line is not provided, we cannot see the exact positions of f and g. However, we can discuss how to write inequalities to compare two integers based on their positions on a number line. If f is located to the left of g on the number line, it means that f is less than g. The inequality for this scenario would be f < g. On the other hand, if f is located to the right of g, then f is greater than g, and the corresponding inequality would be f > g.

You can use an inequality symbol to show how two metric measurements are related. If two numbers are the same, the inequality symbol would be the equal sign, representing they are equivalent. However, without the number line, we cannot determine the exact relationship between f and g, so one must look at the number line to ascertain the correct inequality to use.

Answer:

A

Step-by-step explanation:

Cause I know

Answer: C!!

Step-by-step explanation:

USA Test Prep told me! :)

Answer: The answer is 1. Substitute (2,1) into x+3y=5

Step-by-step explanation: Remember, (2,1) 2 is substituted for x and 1 is used as the y substitute

Answer:

1. 2+3(1)=5

2. 5=5 is true

3. 1=-(2)+3

4. 1=1 is true

x less-than-or-equal-to 6 is the inequality which is represented by the graph.

The problem can be solved by following steps

It is given that

According to above statement a graph is drawn below

The shaded part of the left side describes the value of x which are less than 6 or equal to 6

Hence , The inequality that is represented in the graph is x is less than or equal to 6

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Answer: B

Step-by-step explanatii got it right on edge

Answer:

[tex]\large \boxed{z = \dfrac{83}{6 - C - T}}[/tex]

Step-by-step explanation:

[tex]\begin{array}{rccl}-Cz + 6z & = & Tz + 83 & \\6z -Cz - Tz &= & 83 & \text{Subtracted Tz from each side}\\z(6 - C - T) & = & 83 & \text{Removed the common factor}\\z& = & \mathbf{\dfrac{83}{6 - C - T}} &\text{Divided each side by 6 - C - T}\\\end{array}\\\\\large \boxed{\mathbf{z = \dfrac{83}{6 - C - T}}}[/tex]

Answer:

The slope is -1

Step-by-step explanation:

Let's find the slope between your two points.

(1,4);(6,−1)

(x1,y1)=(1,4)

(x2,y2)=(6,−1)

Use the slope formula:

m= y2−y1/x2−x1  = −1−4/6−1

= −5/5

= −1

Hope this is a better explanation :)

Hope it helps!!!!!!!!!

Answer:

The range of the p-value is: 0.050 < p-value < 0.100.

Step-by-step explanation:

For checking the equivalence of two population variances of independent samples, we use the f-test.

The test statistic is given by:

[tex]F=\frac{S_{1}^{2}}{S_{2}^{2}}\sim F_{\alpha, (n_{1}-1)(n_{2}-1)}[/tex]

It is provided that the hypothesis test is one-tailed.

The computed value of the test statistic is:

F = 4.23.

The degrees of freedom of the numerator and denominator are:

[tex]df_{1}=4\\df_{2}=5[/tex]

Use MS-Excel to compute the p-value as follows:

Step 1: Select function fX → F.DIST.RT.

Step 2: A dialog box will open. Enter the values of f-statistic and the two degrees of freedom.

*See the attachment below.

Step 3: Press OK.

The p-value is, 0.0728.

The range of the p-value is:

0.050 < p-value < 0.100

The value is 50

A fraction represents a part of a whole or, more generally, any number of equal parts.

Given:

p/4 + pq

As, p= 8, q= 6

We have,

8/4 + 8*6

= 2 + 48

= 50

Hence, p/4 + pq= 50

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The equation results to get 50.

To evaluate the expression p/4 + pq when p = 8 and q = 6, follow these steps:

Therefore, the value of the expression when p = 8 and q = 6 is 50.

The complete question is:

Evaluate p/4 +pq when p=8 and q=6

Answer:

4

Step-by-step explanation:

AC = AE = 4 cm

Answer:

The correct answer is (E).

This is not an experiment because no treatment is being imposed upon the customers. Additionally, this study used a stratified sample because independent random samples were selected from two distinct populations of customers.

Step-by-step explanation:

The correct answer is (E) An observational study using a stratified sample.

Stratified sampling is also known as stratified random sampling. The stratified sampling process starts with researchers dividing a diverse population into relatively homogeneous groups called strata, the plural of stratum. Then, they draw a random sample from each group (stratum) and combine them to form their complete representative sample.

Given that a data of a survey, the study found that the mean satisfaction rating was significantly greater among customers that purchased computers that offer tech support.

We need to find which is the best description of this study,

This study used a stratified sample because independent random samples were selected from two distinct populations of customers.

This is not an experiment because no treatment is being imposed upon the customers.

Hence, the best description is an observational study using a stratified sample.

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The complete question is attached.

Answer:

At first we should know that:

The properties of the square are:

The properties of Rhombus

So, Wade is incorrect because the quadrilateral may be Rhombus

And the quadrilateral to be a square, she needs to show that It has two pairs of perpendicular lines.

Answer:

10

3.4641

2.8965  

12.8965  

Step-by-step explanation:

Given: 10, 8, 12, 15, 13, 11, 6, 5  

c = 95%

a. The point estimate of the population mean is the sample mean. The mean is the sum of all values divided by the number of values:

x = 10 + 8 + 12 + 15 + 13 + 11 + 6 + 5 /8

  = 80/8

  = 10

b. The point estimate of the population standard deviation is the sample standard deviation. The variance is the sum of squared deviations from the mean divided by n - 1. The standard deviation is the square root of the variance:  

s = /(10 – 10)^2 +.... + (5– 10)^2/8 – 1  

s = 3.4641

c. Determine the t-value by looking in the row starting with degrees of freedom df = n-1 = 8 –1 = 7 and in the column with [tex]\alpha[/tex] = (1 – c)/2 = 0.025 in table :  

t_[tex]\alpha[/tex]/2 = 2.365  

The margin of error is then:  

E = t_[tex]\alpha[/tex]/2 * s/√n

  = 2.365 x s 3.4641/ √8

  = 2.8965  

d. The confidence intent)] then becomes:  

7.1035 = 10 – 2.8965 = x – E <u<x +E= 10 + 2.8965 = 12.8965  

The point estimate of the population mean will be 10.

The point estimate of the population mean will be calculated thus:

= (10 + 8 + 12 + 15 + 13 + 11 + 11 + 6 + 5) / 8

= 80/8

= 10

Also, the margin of error will be:

= 2.365 × 3.4641/✓8

= 2.8965

In conclusion, the margin of error is 2.8965.

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Answer:

For this case we want to test if there is a higher incidence of smoking among women than among men in a neighborhood (alternative hypothesis). And we define p1 for men and p2 for women, so for this case the best system of hypothesis are:

Null hypothesis:[tex] p_1- p_2 \geq 0[/tex]

Alternative hypothesis: [tex]p_1 -p_2 <0 [/tex]

And the best option would be:

a) h0: p1-p2 ≥ 0 h1: p1-p2 < 0

Step-by-step explanation:

Previous concepts

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".  

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".  

Solution to the problem

For this case we want to test if there is a higher incidence of smoking among women than among men in a neighborhood (alternative hypothesis). And we define p1 for men and p2 for women, so for this case the best system of hypothesis are:

Null hypothesis:[tex] p_1- p_2 \geq 0[/tex]

Alternative hypothesis: [tex]p_1 -p_2 <0 [/tex]

And the best option would be:

a) h0: p1-p2 ≥ 0 h1: p1-p2 < 0

Answer:

7.64% probability that they spend less than $160 on back-to-college electronics

Step-by-step explanation:

Problems of normally distributed samples can be 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 = 237, \sigma = 54[/tex]

Probability that they spend less than $160 on back-to-college electronics

This is the pvalue of Z when X = 160. So

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

[tex]Z = \frac{160 - 237}{54}[/tex]

[tex]Z = -1.43[/tex]

[tex]Z = -1.43[/tex] has a pvalue of 0.0763

7.64% probability that they spend less than $160 on back-to-college electronics