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DanielleElmas [232]
3 years ago
10

In 2016 there were about 64,500 cinema screens in a country. If about 29.9% of screens in the country were digital 3-D screens,

find the approximate number of digital 3-D screens.
Mathematics
1 answer:
zhenek [66]3 years ago
6 0

Answer:

Step-by-step explanation:

In 2016 there were about 64,500 cinema screens in a country.

If about 29.9% of screens in the country were digital 3-D screens, we would determine the approximate number of 3D screens by dividing the Percentage of 3D screens by the total percentage of 100 and multiplying by the total number of cinema screens. It becomes

29.9/ 100 × 64500 = 0.299 × 64500

= 19285.5

The number of 3D screens is approximately 19286

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A given population proportion is .25. What is the probability of getting each of the following sample proportions
anyanavicka [17]

This question is incomplete, the complete question is;

A given population proportion is .25. What is the probability of getting each of the following sample proportions

a) n = 110 and = p^ ≤ 0.21, prob = ?

b) n = 33 and p^ > 0.24, prob = ?

Round all z values to 2 decimal places. Round all intermediate calculation and answers to 4 decimal places.)

Answer:

a) the probability of getting the sample proportion is 0.1660

b) the probability of getting the sample proportion is 0.5517

Step-by-step explanation:

Given the data in the questions

a)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 110

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 110 ) √( 0.1875 / 110 ) = 0.0413

Now, P( p^ ≤ 0.21 )

= P[ (( p^ - μ ) /S.D) < (( 0.21 - μ ) / S.D)

= P[ Z < ( 0.21 - 0.25 ) / 0.0413)

= P[ Z < -0.04 / 0.0413]

= P[ Z < -0.97 ]

from z-score table

P( X ≤ 0.21 ) = 0.1660

Therefore, the probability of getting the sample proportion is 0.1660

b)

population proportion = 0.25

q = 1 - p = 1 - 0.25 = 0.75

sample size n = 33

mean = μ = 0.25

S.D = √( p( 1 - p) / n ) = √(0.25( 1 - 0.25) / 33 ) = √( 0.1875 / 33 ) = 0.0754

Now, P( p^ > 0.24 )  

= P[ (( p^ - μ ) /S.D) > (( 0.24 - μ ) / S.D)

= P[ Z > ( 0.24 - 0.25 ) / 0.0754 )

= P[ Z > -0.01 / 0.0754  ]

= P[ Z > -0.13 ]

= 1 - P[ Z < -0.13 ]

from z-score table

{P[ Z < -0.13 ] = 0.4483}

1 - 0.4483

P( p^ > 0.24 )  = 0.5517

Therefore, the probability of getting the sample proportion is 0.5517

6 0
3 years ago
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