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Mandarinka [93]
3 years ago
14

Suppose a certain scale is not calibrated correctly, and as a result, the mass of any object is displayed as 0. 75 kilogram less

than its actual mass. What is the correlation between the actual masses of a set of objects and the respective masses of the same set of objects displayed by the scale?
SAT
1 answer:
r-ruslan [8.4K]3 years ago
4 0

The correlation between the actual masses of a set of objects and the respective masses of the same set of objects displayed by the scale is; 1

<h3>What is the Correct Scale Calibration?</h3>

We are told that a certain scale is not calibrated correctly,

Mass displayed = 0.75 kg (less than its actual mass).

The scale of display = 1

Thus, the correlation between actual mass and displayed mass is 1 due to the fact that correlation is independent of change of origin and scale and correlation between any value and value minus 0.75 will be one.

Thus, the correlation between the actual masses of a set of objects and the respective masses of the same set of objects displayed by the scale is 1.

Read more about Scale Calibration at; brainly.com/question/13853335

#SPJ1

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5 0
2 years ago
In a recent year, about 22% of Americans 18 years and older are single. What is the probability that in a random sample of 200 A
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Using the normal approximation to the binomial, it is found that there is a 0.0107 = 1.07% probability that more than 30 are single.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
  • The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with \mu = np, \sigma = \sqrt{np(1-p)}.

In this problem, the proportion and the sample size are, respectively, p = 0.22 and n = 200, hence:

\mu = np = 200(0.22) = 44

\sigma = \sqrt{np(1 - p)} = \sqrt{200(0.22)(0.78)} = 5.8583

The probability that more than 30 are single, using continuity correction, is P(X > 30.5), which is <u>1 subtracted by the p-value of Z when X = 30.5</u>, hence:

Z = \frac{X - \mu}{\sigma}

Z = \frac{30.5 - 44}{5.8583}

Z = -2.3

Z = -2.3 has a p-value of 0.0107.

0.0107 = 1.07% probability that more than 30 are single.

More can be learned about the normal distribution at brainly.com/question/24663213

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