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Amanda [17]
3 years ago
14

A frequency table is shown for the height's of 6th graders in a class at Oakville Elementary School. What is the probability tha

t a sixth grader in this class is 52 inches tall?
A) 0.14
B) 0.21
C) 0.25
D) 0.28

Mathematics
2 answers:
const2013 [10]3 years ago
7 0

Answer:

<h2>B) 0.21</h2>

Step-by-step explanation:

To find the probability, we first need to know the total number of students, which is the sum of all frequencies

N=5+2+0+3+6+5+4+2+0+1=28

Then, we look for the total number of students 52 inches tall.

According to the table, there are only 6 students 52 inches tall.

So, the probability would be the quotient between 6 and 28

P=\frac{6}{28} \approx 0.21

Therefore, the right answer is B.

Xelga [282]3 years ago
6 0
The answer would be B) .21
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Step-by-step explanation:

So if 8 pens cost $4.72, one pen would cost $0.59. 0.59 x 13 will be equal to 7.67.

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Step-by-step explanation:

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3 years ago
Select the correct answer. Shape 1 and shape 2 are plotted on a coordinate plane. Which statement about the shapes is true? A. S
IceJOKER [234]

Answer:

Since I cant say which answer due to no graph, I'll tell you How to do so.

Step-by-step explanation:

if it is A, then the there is at least one angle or line length that is not the same. To find the area of a grided shape, use the traingle theorm of a^2+b^2=c^2.

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6 0
3 years ago
A new roller coaster at an amusement park requires individuals to be at least​ 4' 8" ​(56 ​inches) tall to ride. It is estimated
Maksim231197 [3]

Answer:

a) 34.46% of​ 10-year-old boys is tall enough to ride this​ coaster.

b) 78.81% of​ 10-year-old boys is tall enough to ride this​ coaster

c) 44.35% of​ 10-year-old boys is tall enough to ride the coaster in part b but not tall enough to ride the coaster in part​ a

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

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 question, we have that:

\mu = 54, \sigma = 5

a. What proportion of​ 10-year-old boys is tall enough to ride the​ coaster?

This is 1 subtracted by the pvalue of Z when X = 56.

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{56 - 54}{5}

Z = 0.4

Z = 0.4 has a pvalue of 0.6554

1 - 0.6554 = 0.3446

34.46% of​ 10-year-old boys is tall enough to ride this​ coaster.

b. A smaller coaster has a height requirement of 50 inches to ride. What proportion of​ 10-year-old boys is tall enough to ride this​ coaster?

This is 1 subtracted by the pvalue of Z when X = 50.

Z = \frac{X - \mu}{\sigma}

Z = \frac{50 - 54}{5}

Z = -0.8

Z = -0.8 has a pvalue of 0.2119

1 - 0.2119 = 0.7881

78.81% of​ 10-year-old boys is tall enough to ride this​ coaster.

c. What proportion of​ 10-year-old boys is tall enough to ride the coaster in part b but not tall enough to ride the coaster in part​ a?

Between 50 and 56 inches, which is the pvalue of Z when X = 56 subtracted by the pvalue of Z when X = 50.

From a), when X = 56, Z has a pvalue of 0.6554

From b), when X = 50, Z has a pvalue of 0.2119

0.6554 - 0.2119 = 0.4435

44.35% of​ 10-year-old boys is tall enough to ride the coaster in part b but not tall enough to ride the coaster in part​ a

5 0
3 years ago
The author drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 199 times. Here are the observed
Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28​, 29​, 47​, 40​, 22​, 33.

Probability of an Event

      =\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911

Result is not significant.

3.

(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378

Result is not significant.

4.

(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349

Result is not significant.

5.

(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773

Result is not significant.

6.

(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair​ die.

   

8 0
3 years ago
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