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IceJOKER [234]
3 years ago
14

In all, 1000 students took a statistics exam worth 150 points. The first quartile (or Q1) for all 1000 scores is 90 points. Abou

t how many students had scores greater than 90 points?
Mathematics
2 answers:
nasty-shy [4]3 years ago
5 0

Answer:

i think 750

Step-by-step explanation:

Brums [2.3K]3 years ago
4 0

Answer:

The number of students who scored more than 90 points is 750.

Step-by-step explanation:

Quartiles are statistical measures that the divide the data into four groups.

The first quartile (Q₁) indicates that 25% of the observation are less than or equal to Q₁.

The second quartile (Q₂) indicates that 50% of the observation are less than or equal to Q₂.

The third quartile (Q₃) indicates that 75% of the observation are less than or equal to Q₃.

It is provided that the first quartile is at 90 points.

That is, P (X ≤ 90) = 0.25.

The probability that a student scores more than 90 points is:

P (X > 90) = 1 - P (X ≤ 90)

                = 1 - 0.25

                = 0.75

The number of students who scored more than 90 points is: 1000 × 0.75 = 750.

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Which expression is equivalent to 2^3/2^-8
Mkey [24]

Answer:

2^{11}

Step-by-step explanation:

<u>Key indices rule: When dividing powers, subtract the indices and keep the base numbers constant.</u>

<u />\frac{2^{3} }{2^{-8} } Based on the rule above, the indices are determined by 3--8, which is 11.

The base number is kept constant.

Therefore, the answer is 2^{11}.

7 0
3 years ago
Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa
Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation  

n=580 represent the random sample taken    

X represent the seafood sold in the country that is mislabeled or misidentified by the people

\hat p=0.25 estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

\alpha=0.01 represent the significance level (no given, but is assumed)    

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204

0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

Part c

A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

4 0
4 years ago
A map uses a scale of 1 in :20mi. Find the the actual distance of 5.5 inches on the map. help me​
astraxan [27]

Answer:

110

Step-by-step explanation:

5.5 x 20 = 110

4 0
3 years ago
Decide whether the function is an exponential growth or exponential decay function, and find the constant percentage rate of gro
Norma-Jean [14]
Exponential growth function; 7%
5 0
4 years ago
Read 2 more answers
Sarah has gone four weeks. Micah has been recording the daily high temperature. During that time, the high temperature has been
pickupchik [31]
The experimental probability that the high temp will be below 45 F on the 29th day is, 4.5 out of 29 days, that is because 20 of the days have had lower temperatures, so they left 9 days of and from those 9 days, they could be lower or greater that 45F (50% each side), that is:
4.5/29 = 0.155 = 15.5%
that is the exp probability
8 0
4 years ago
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