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4 years ago

The length of a movie falls on a normal distribution. About 95% of movies fall between 75 minutes and 163 minutes.

Mathematics

1 answer:

ioda4 years ago

7 0

Answer:

a) x = 119 minutes

the mean value for average movie length in minutes is 119 minutes

b) margin of error M.E = 44 minutes

Note; Since the number of samples used is not given, the standard deviation r cannot be calculated using the equation

M.E = zr/√n

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

x+/-zr/√n

x+/-M.E

M.E = zr/√n

Given that;

M.E = margin of error

Mean = x

Standard deviation = r

Number of samples = n

Confidence interval = 95%

z value (at 95% confidence) = 1.96

a) The mean value x can be calculated as;

x = (a+b)/2

Where a and b are the lower and upper bounds of the confidence interval;

a = 75 minutes

b = 163 minutes

substituting the values;

x = (75+163)/2

x = 119 minutes

the mean value for average movie length in minutes is 119 minutes

b) the margin of error M.E can be calculated as;

M.E = (b-a)/2

Substituting the values;

M.E = (163-75)/2

M.E = 44 minutes

Since the number of samples used is not given, the standard deviation r cannot be calculated using the equation

M.E = zr/√n

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It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

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If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

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$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

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