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Anna11 [10]
3 years ago
8

At the school carnival, the sixth graders are making directional arrows, as shown below.

Mathematics
1 answer:
Citrus2011 [14]3 years ago
6 0

Answer:

They would need A. 26 square inches of red paint. I hope this helped

Step-by-step explanation:

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The GMAC Insurance company reported that the mean score on the National Drivers Test was 70.5 with a standard deviation of 3.6 p
aliina [53]

In statistics, the 68–95–99.7 rule, also known as the three-sigma rule or empirical rule, states that nearly all values lie within three standard deviations of the mean in a normal distribution.

About 68.27% of the values lie within one standard deviation of the mean. Similarly, about 95.45% of the values lie within two standard deviations of the mean. Nearly all (99.73%) of the values lie within three standard deviations of the mean.

In mathematical notation, these facts can be expressed as follows, where x is an observation from a normally distributed random variable, μ is the mean of the distribution, and σ is its standard deviation:

P( μ – σ ≤ x ≤ μ + σ ) ≈ 0.6827

P( μ –2σ ≤ x ≤ μ + 2σ ) ≈ 0.9545

P( μ –3σ ≤ x ≤ μ + 3σ ) ≈ 0.9973

So here A = μ – σ = 78.7 – 3.5 = 75.2 (answer

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4 years ago
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Step-by-step explanation:

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3 years ago
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Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
user100 [1]

Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

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

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

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

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

5 0
3 years ago
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