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2 years ago

Find the values of the mode when median is given to be 5 and mean is 7.

Mathematics

2 answers:

Reil [10]2 years ago

5 0

Answer:

Mode = 1

Step-by-step explanation:

Relation between the Central Measures of Tendency

Mean, Median, and Mode are commonly referred to as the Central Measures of Tendency
The formula between the three is given by :
⇒ Mode = 3Median - 2Mean or Mode + 2Mean = 3Median

Solving

We know that :

Median = 5
Mean = 7

Therefore,

Mode = 3(5) - 2(7)
Mode = 15 - 14
Mode = 1

Ugo [173]2 years ago

3 0

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

We know the relation between mean, Median and mode. that is :

$\qquad \sf \dashrightarrow \:mode = 3 \: median - 2 \: mean$

now, plug in the values ~

$\qquad \sf \dashrightarrow \:mode = 3(5) - 2(7)$

$\qquad \sf \dashrightarrow \:mode = 15 - 14$

$\qquad \sf \dashrightarrow \:mode = 1$

Hence, value of mode is 1

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In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a samp

dsp73

Answer:

$(0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401$

$(0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380$

We are confident at 95% that the difference between the two proportions is between $-0.4401 \leq p_B -p_A \leq -0.1380$

1. -.4401 ≤ p1 - p2 ≤ -.1380

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

Step-by-step explanation:

In San Jose a sample of 73 mail carriers showed that 30 had been bitten by an animal during one week. In San Francisco in a sample of 80 mail carriers, 56 had received animal bites. Is there a significant difference in the proportions? Use a 0.05. Find the 95% confidence interval for the difference of the two proportions. Sellect all correct statements below based on the data given in this problem.

1. -.4401 ≤ p1 - p2 ≤ -.1380

2. -.4401 ≤ p1 - p2 ≤ .1380

3. The rate of mail carriers being bitten in San Jose is statistically greater than the rate San Francisco at α = 5%

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

5. The rate of mail carriers being bitten in San Jose and San Francisco are statistically equal at α = 5%

Solution to the problem

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$p_1$ represent the real population proportion for San Jose

$\hat p_1 =\frac{30}{73}=0.411$ represent the estimated proportion for San Jos

$n_1=73$ is the sample size required for San Jose

$p_2$ represent the real population proportion for San Francisco

$\hat p_2 =\frac{56}{80}=0.7$ represent the estimated proportion for San Francisco

$n_2=80$ is the sample size required for San Francisco

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval for the difference of two proportions would be given by this formula

$(\hat p_1 -\hat p_1) \pm z_{\alpha/2} \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1} +\frac{\hat p_2 (1-\hat p_2)}{n_2}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$(0.411-0.7) - 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.4401$

$(0.411-0.7) + 1.96 \sqrt{\frac{0.411(1-0.411)}{73} +\frac{0.7(1-0.7)}{80}}=-0.1380$

We are confident at 95% that the difference between the two proportions is between $-0.4401 \leq p_B -p_A \leq -0.1380$

Since the confidence interval contains all negative values we can conclude that the proportion for San Jose is significantly lower than the proportion for San Francisco at 5% level.

Based on this the correct options are:

1. -.4401 ≤ p1 - p2 ≤ -.1380

4. The rate of mail carriers being bitten in San Jose is statistically less than the rate San Francisco at α = 5%

8 0

2 years ago

Find the y-intercept of the following equation. Simplify your answer. 4x + y = 7

Bogdan [553]

4x + y = 7
y = -4x + 7

using y = mx + b

b being your y intercept, (0,7)

5 0

2 years ago

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