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IrinaVladis
7 days ago
10

9. Apply The number of people in the United States who are at least 100

Mathematics
1 answer:
zzz [9K]7 days ago
8 0
In 1980, if there were N individuals aged 100 or older, then by 2010, the number grew to N*1.66 for those 100 and above. A straightforward conditional expression can be framed as: If P, then Q, where P represents the hypothesis and Q is the conclusion. We understand that "The count of individuals aged 100 years or older increased by about 66% from 1980 to 2010," meaning that if we had N individuals aged 100 or older in 1980, we will have N*(166%/100%) = N*1.66 in 2010, allowing us to write a conditional statement: If there were N individuals aged 100 years old in 1980, then by 2010, we had N*1.66 individuals who were at least 100 years old.
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The average lifetime of a light bulb is 3,000 hours with a standard deviation of 696 hours. A simple random sample of 36 bulbs i
Leona [9260]

Answer:

Answer and Explanation:

We have:

Population mean,

μ

=

3

,

000

hours

Population standard deviation,

σ

=

696

hours

Sample size,

n

=

36

1) The standard deviation for the sampling distribution:

σ

¯

x

=

σ

√

n

=

696

√

36

=

116

2) By the central limit theorem, the sampling distribution's expected value matches the population mean.

Thus:

The expected value of the sampling distribution equals the population mean,

μ

¯

x

=

μ

=

3

,

000

The standard deviation of the sampling distribution,

σ

¯

x

=

116

The sampling distribution of

¯

x

is roughly normal due to a sample size greater than

30

.

3) The likelihood that the average lifespan of the sample falls between

2670.56

and

2809.76

hours:

P

(

2670.56

<

x

<

2809.76

)

=

P

(

2670.56

−

3000

116

<

z

<

2809.76

−

3000

116

)

=

P

(

−

2.84

<

z

<

−

1.64

)

=

P

(

z

<

−

1.64

)

−

P

(

z

<

−

2.84

)

=

0.0482

In Excel: =NORMSDIST(-1.64)-NORMSDIST(-2.84)

4) The probability of the average life in the sample exceeding

3219.24

hours:

P

(

x

>

3219.24

)

=

P

(

z

>

3219.24

−

3000

116

)

=

P

(

z

>

1.89

)

=

0.0294

In Excel: =NORMSDIST(-1.89)

5) The likelihood that the sample's average life is lower than

3180.96

hours:

P

(

x

<

3180.96

)

=

P

(

z

<

3180.96

−

3000

116

)

=

P

(

z

<

1.56

)

=

0.9406

6 0
17 days ago
1) Jodi liked to collect stamps. On 3 different days she bought 6 stomps. Then she
babunello [8402]

Answer:

4

Step-by-step explanation:

6-4+2*5=12

12/3

4

6 0
6 days ago
Two race cars,car x an y,are at the starting point of a two km track at the same time.car x and car y make one lap every 60 s an
AnnZ [9071]

The answer is in the image provided.

Please refer to the accompanying image.

4 0
27 days ago
Mike runs for the president of the student government and is interested to know whether the proportion of the student body in fa
zzz [9066]

Answer:

We conclude that less than or equal to 50% of the student body supports him significantly.

Step-by-step explanation:

Mike, while campaigning for the student government presidency, is keen to determine if more than 50% of the student body supports him.

A random sample of 100 students was surveyed, with 55 expressing support for Mike.

Let p = the proportion of students backing Mike.

Thus, Null Hypothesis, H_0: p \leq 50% {indicating that the proportion of supporters among the student body is significantly less than or equal to 50%}

Alternate Hypothesis, H_A: p > 50% {indicating that the proportion of supporters among the student body is significantly more than 50%}

The test statistics to be utilized here One-sample z proportion statistics;

T.S. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } ~ N(0,1)

where, \hat p = the sampled proportion in favor of Mike = \frac{55}{100} = 0.55

n = number of sampled students = 100

Therefore, test statistics = \frac{0.55-0.50}{\sqrt{\frac{0.55(1-0.55)}{100} } }

= 1.01

The z test statistic is 1.01.

At a significance level of 0.05, the z table provides a critical value of 1.645 for a right-tailed test.

Given that our test statistic falls below the critical value of z (1.01 < 1.645), we lack sufficient evidence to reject the null hypothesis as it remains outside the rejection region, leading to failure to reject our null hypothesis.

As a result, we conclude that less than or equal to 50% of the student body is in favor of him or the proportion supporting Mike is not significantly greater than 50%.

4 0
25 days ago
In Jane’s calculus class at a large university, the final exam had a mean score of 75 and a standard deviation of 10. In Peter’s
zzz [9066]

Response: 100

Detailed breakdown:

5 0
6 days ago
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