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valentina_108 [34]
3 years ago
10

Use the following information to answer the question. The mean age of lead actresses from the top ten grossing movies of 2010 wa

s 29.6 years with a standard deviation of 6.35 years. Assume the distribution of the actresses' ages is approximately unimodal and symmetric. Between what two values would you expect to find about 95% of the lead actresses ages
Mathematics
1 answer:
Alex787 [66]3 years ago
4 0

Answer:

You should expect to find about 95% of the lead actresses ages between 16.9 years and 42.3 years.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 29.6 years, Standard deviation = 6.35 years.

Between what two values would you expect to find about 95% of the lead actresses ages

By the Empirical Rule, within 2 standard deviations of the mean. So

29.6 - 2*6.35 = 16.9 years

29.6 + 2*6.35 = 42.3 years

You should expect to find about 95% of the lead actresses ages between 16.9 years and 42.3 years.

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At a certain high school, 15 students play stringed instruments and
Aleks [24]

Answer:

<em>0</em> is the probability that a  randomly selected student plays both a stringed and a brass  instrument.

Step-by-step explanation:

Given that:

Number of students who play stringed instruments, N(A) = 15

Number of students who play brass instruments, N(B) = 20

Number of students who play neither, N(A \cup B)' = 5

<u>To find:</u>

The probability that a randomly selected students plays both = ?

<u>Solution:</u>

Total Number of students = N(A)+N(B)+N(A \cup B)' =15 + 20 + 5 = 40

(As there is no student common in both the instruments we can simply add the three values to find the total number of students)

As per the venn diagram, no student plays both the instruments i.e.

N(A\cap B) =0

Formula for probability of an event E can be observed as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

P(A\cap B) = \dfrac{N(A\cap B)}{N(U)}\\\Rightarrow \dfrac{0}{40}\\\Rightarrow P(A\cap B) = 0

So, <em>0</em> is the probability that a  randomly selected student plays both a stringed and a brass  instrument.

5 0
3 years ago
Find the mode in 18 7 4 19 3 18 19 12
scoundrel [369]
Hello there,

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18 is your answer.

~Jurgen
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Jlenok [28]
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