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Mars2501 [29]
3 years ago
14

Lamar wants to determine the movie preferences

Mathematics
2 answers:
shepuryov [24]3 years ago
8 0

Answer:

male comedy: 31 male drama: 9

female comedy: 36 female drama: 24

Step-by-step explanation:

add f+m=100 total

31 males for comedy

40 males - 31 males = 9 males for drama

55% chose comedy so 55-31 males = 24 females for comedy

60 total females - 24 for comedy = 36 females for drama

adelina 88 [10]3 years ago
8 0

Answer:

                         Dramas              Comedies

Males                   9                             31

Females             36                            24

Step-by-step explanation:

It is given that Lamar surveyed 40 males and 60 females about whether they prefer dramas or comedies.

Total males = 40

Total females = 60

Total people surveyed = 40 + 60 = 100

It is given that 55% of surveyed people preferred comedies.

People preferred comedies = 100\times \frac{55}{100}=55

It is given that males who preferred comedies = 31

Males who preferred drama is the difference of number of males and males who preferred comedies.

Males, drama = 40 - 31 = 9

Females who preferred comedies is the difference of people who preferred comedies and males who preferred comedies.

Females, comedies = 55 - 31 = 24

Females who preferred drama is the difference of number of females and females who preferred comedies.

Females, drama = 60 - 24 = 36

Therefore the table can be filled as

                         Dramas              Comedies

Males                   9                             31

Females             36                            24

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Answer:

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Step-by-step explanation:

From the question,

A map has a scale of 1: 1250000.

We are told that:

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ki77a [65]

Answer:

a)

i.  x=- - - - - -

ii.  o=- - - - - -

iii.  Sx=- - - - -

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So, the distribution is  

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e)

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iii. The error bound is 0.0411

f)

i. The 98% confidence interval for the population mean weight of the candies is  1.9418 , 2.0582

iii. The error bound is 0.0582

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Step-by-step explanation:

a)

i.  x=- - - - - -

ii.  o=- - - - - -

iii.  Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.  

c) The statistics variable X  is a measure on a sample that is used as an estimate of the population mean.X  is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters  because the population standard deviation is known.

N(X, o/\sqrt{n})

So, the distribution is  

N(2, 0.12/\sqrt{16})

e)

i. The 90% confidence interval for the population mean weight of the candies is  1.9589 , 2.0411

iii. The error bound is 0.0411

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i. The 98% confidence interval for the population mean weight of the candies is  1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part  is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

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slava [35]
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