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4 years ago

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After the data was recorded, Leah realized that she left out a student who handed out 42 flyers. How would the mean number of fl

yers handed out be affected if Leah added this number to the data shown in the histogram?

1 answer:

Yuliya22 [10]4 years ago

8 0

Answer:

Outlier An extreme value in a set of data which is much higher or lower than the other numbers. ... Outliers affect the mean value of the data but have little effect on the median or mode of a given set of data. Your question asks how number 42 would affect the mean, that depends on two things, is the 42 higher or lower than the mean before 42 is added to the data? If the 42 is higher, the mean would increase, if the 42 is lower, the mean would decrease. Hope that helps.

Step-by-step explanation:

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There are 40 grams of sugar in a 12 oz can of Dr Pepper. If your max intake of sugar permitted for the day is 25 grams, how much

Zarrin [17]

To answer this question, we can set up a proportion:

$\frac{40g}{12oz} =\frac{25g}{x}$

Then we cross multiply and solve for x:

$40x=300$

$x=7.5oz$

So you can drink 7.5 oz of the Dr. Pepper in one day.

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4 years ago

Read 2 more answers

15 boys can cut the grass on the football field in 12days. how many days will it take 9 boys

Inessa [10]

X=15-12
X=3(number of days in difference to the number of people)
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3 years ago

Listed below are the number of years it took for a random sample of college students to earn bachelor's degrees (based on data f

schepotkina [342]

Answer:

a) Mean = 6.5, sample standard deviation = 3.50

b) Standard error = 0.7826

c) Point estimate = 6.5

d) Confidence interval: (5.1469 ,7.8531)

Step-by-step explanation:

We are given the following data set for students to earn bachelor's degrees.

4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{130}{20} = 6.5$

Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5

$S.D = \sqrt{\frac{233.5}{19}} = 3.50$

b) Standard Error

$= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826$

c) Point estimate for the mean time required for all college is given by the sample mean.

$\bar{x} = 6.5$

d) 90% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729$

$6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)$

e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.

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4 years ago

A company bought New Year's greeting cards to send to its valued customers. Forty-five of the cards they sent cost $2.40 each. T

ASHA 777 [7]

Answer: $262.80

Step-by-step explanation: 45x2.40 + 36x2.9 + 28x1.8 = 262.8

8 0

3 years ago

Assume that we have m coins. We toss each one of them n times. The probability of heads showing up for each coin isp. What’s the

olga nikolaevna [1]

Answer:

$1-(1-p^n)^m$

Step-by-step explanation:

For a coin, the probability of head showing in a single toss is $p$ .

$P(H)=p$

Its complement, the probability of not head is

$P(\Sim H)=1-p$

This is a binomial distribution. In $n$ tosses, the probability of having all heads (i.e. $n$ heads) is

$P(\text{all heads})=\binom{n}{n}p^n(1-p)^0=p^n$

Let's call this value $a$ .

For $m$ coins, we determine the probability of at least 1 coin showing all heads by first finding its complement i.e. the probability of no coin showing all heads. This is also a binomial distribution.

$P(\text{no coin showing all heads})=\binom{m}{0}a^0(1-a)^m=(1-a)^m$

$P(\text{at least 1 coin showing all heads})=1-P(\text{no coin showing all heads})$

$P(\text{no coin showing all heads})=1-(1-a)^m=1-(1-p^n)^m$

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3 years ago