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sergey [27]
2 years ago
15

Recently a study of fans attending the new york mets baseball games was conducted and 500 fans were surveyed. in forming a frequ

ency distribution of the number of miles fans traveled from home to the stadium, it was found that 247 fans traveled between 0 and 5 miles. based on this information what was the relative frequency for this class? select one:
a. can't be determined without more information.
b. 0.30
c. 0.494
d. 0.247
Mathematics
1 answer:
Misha Larkins [42]2 years ago
8 0

Relative frequency is the value assigned to each class in the proportion of the total data set that belongs in the class.

Relative frequency formula is :

\frac{class-Frequency}{Total- observations}--------------------(1)

So for class 0 to 5 miles we have 247 fans. So this class frequency is 247.

Total fans surveyed is 500

So plug these values in relative frequency formula given by (1)

relative frequency = \frac{247}{500}

On simplifying we get

relative frequency = 0.494

So choice (c) 0.494 is the right answer

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2 years ago
The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.512.512, point, 5 years; the stan
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This question was not written properly

Complete Question

The lifespans of lions in a particular zoo are normally distributed. The average lion lives 12.5 years; the standard deviation is 2.4 years. Use the empirical rule (68-95-99.7\%)(68−95−99.7%) to estimate the probability of a lion living between 5.3 to 10. 1 years.

Answer:

Thehe probability of a lion living between 5.3 to 10. 1 years is 0.1585

Step-by-step explanation:

The empirical rule formula states that:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

3) 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

Mean is given in the question as: 12.5

Standard deviation : 2.4 years

We start by applying the first rule

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

μ - σ

12.5 -2.4

= 10.1

We apply the second rule

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

μ – 2σ

12.5 - 2 × 2.4

12.5 - 4.8

= 7.7

We apply the third rule

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

μ - 3σ

= 12.5 - 3(2.4)

= 12.5 - 7.2

= 5.3

From the above calculation , we can see that

5.3 years corresponds to one side of 99.7%

Hence,

100 - 99.7%/2 = 0.3%/2

= 0.15%

And 10.1 years corresponds to one side of 68%

Hence

100 - 68%/2 = 32%/2 = 16%

So,the percentage of a lion living between 5.3 to 10. 1 years is calculated as 16% - 0.15%

= 15.85%

Therefore, the probability of a lion living between 5.3 to 10. 1 years

is converted to decimal =

= 15.85/ 100

= 0.1585

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<h3>Answer with explanation:</h3>

It is given that:

Circle 1 has center (−4, −7) and a radius of 12 cm.

Circle 2 has center (3, 4) and a radius of 15 cm.

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