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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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Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent
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Answer:

0.4

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Given

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This question can be Solved by using Venn diagram

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7 0
1 year ago
For questions 2-5, the number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 a
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Answer:

a. -1.60377

b. 0.25451

c. 0.344

d. Option b) 78th

Step-by-step explanation:

The number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 and a standard deviation of 2.12.

a)What is the z-score value of a randomly selected bag of Skittles that has 35 Skittles? a) 1.62 b) -1.62 c) 3.40 d) -3.40 e)1.303.

The formula for calculating a z-score is is z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

z = 35 - 38.4/2.12

= -1.60377

Option b) -1.62 is correct

b) What is the probability that a randomly selected bag of Skittles has at least 37 Skittles? a) .152 b) .247 c) .253 d).747e).7534. .

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

z = (37 - 38.4)/2.12

= -0.66038

P-value from Z-Table:

P(x<37) = 0.25451

The probability that a randomly selected bag of Skittles has at least 37 Skittles is 0.25451

Option c) .253 is.correct

c) What is the probability that a randomly selected bag of Skittles has between 39 and 42 Skittles? a) .112 b) .232 c) .344 d).457 e).6125.

z = (x-μ)/σ

Mean of 38.4 and a standard deviation of 2.12.

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z = (39 - 38.4)/2.12

= 0.28302

Probability value from Z-Table:

P(x = 39) = 0.61142

For 42 Skittles

z = (42 - 38.4)/2.12

= 1.69811

Probability value from Z-Table:

P(x = 42) = 0.95526

The probability that a randomly selected bag of Skittles has between 39 and 42 Skittles is:

P(x = 42) - P(x = 39

0.95526 - 0.61142

0.34384

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Option c is.correct

d) What is the percentile rank of a randomly selected bag of Skittles that has 40 Skittles in it? a)82nd b) 78th c) 75th d)25th e)22nd

z = (x-μ)/σ

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z = (40 - 38.4)/2.12

= 0.75472

P-value from Z-Table:

P(x = 40) = 0.77479

Converting to percentage = 0.77479× 100

= 77. 479%

≈ 77.5

Percentile rank = 78th

7 0
2 years ago
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5 0
2 years ago
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