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2 years ago

8

The graph shows the quantity of each piece in a toy building kit. One piece is missing. What is the probability that the missing

piece is a red piece with four holes?

1 answer:

Ivahew [28]2 years ago

8 0

Answer:

2/35

Step-by-step explanation:

The only way we can solve this problem is to assume that the given complement of pieces is of a full kit before one went missing. The total number of pieces in the kit is ...

... 9 + 8 + 12 + 6 = 35

Of those, 2 are red with four holes. Thus the probability that a randomly chosen piece is red with 4 holes is 2 out of 35, or 2/35.

____

If this is the kit contents after the piece went missing, we have no way of knowing the color or hole count of the missing piece. If the kit is supposed to contain 13 blue pieces, for example, then the probability is zero that the missing piece is red.

You might be interested in

Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se

Ugo [173]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

$n = 175$

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that $p = 0.467$ . So

$\mu = E(X) = np = 175*0.467 = 81.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6$

This probability, using continuity correction, is $P(X \geq 85 - 0.5) = P(X \geq 84.5)$ , which is 1 subtracted by the pvalue of Z when X = 84.5. So

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

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

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

$Z = \frac{89.5 - 81.725}{6.6}$

$Z = 1.18$

$Z = 1.18$ has a pvalue of 0.8810.

X = 79.5

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

$Z = \frac{79.5 - 81.725}{6.6}$

$Z = -0.34$

$Z = -0.34$ has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that $p = 0.063$

$\mu = E(X) = np = 175*0.063 = 11.025$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141$

This probability, using continuity correction, is $P(X < 12 - 0.5) = P(X < 11.5)$ , which is the pvalue of Z when X = 11.5. So

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

$Z = \frac{11.5 - 11.025}{3.2141}$

$Z = 0.15$

$Z = 0.15$ has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that $p = 0.227$

$\mu = E(X) = np = 175*0.227 = 39.725$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

Using continuity correction, this probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

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

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

8 0

2 years ago

An insurance adjuster processes the claims of six policy holders in an eight hour work day. The adjuster uses $5 in gasoline and

Bad White [126]

Answer:

D

Step-by-step explanation:

Labor productivity is is measured per hours.

Hence, the work he does [amount of claims] will be distributed amongst the amount of hours he works. The amount of dollar he uses <em>doesn't matter for labor productivity.</em>

<em />

Thus, his work is 6 claims. He does that is 8 hours. So claim/hr is:

6/8 = 0.75 claims per hour

Answer choice D is right.

6 0

2 years ago

You are giving the following amounts: $190,258.50; $152,698.00; $122,753.00; $220,523.00; $231,951.00. What is the average of th

blondinia [14]

Answer:

Hey!

The average of these amounts is : $183,636.70!‬

Step-by-step explanation:

To find the average, we have to add all the values up which gives...

= 918,183.50!

And then divide this number by the amount of values you added up...

*918,183.50 divided by 5 gives = $183,636.70*

So your average is $183,636.70!!

HOPE THIS HELPS!!‬‬

4 0

2 years ago

a guy wire makes a 67 degree angle with the ground. walking out 32 ft further grom the tower,the angle of elevation to the top o

Shalnov [3]

Answer:

39.5 ft

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you of the relation between angles and sides of a right triangle.

Tan = Opposite/Adjacent

This lets us write two equations in two unknowns:

tan(67°) = AD/CD . . . . . . . . . . angle at guy point

tan(39°) = AD/(CD+32) . . . . . .angle 32' farther

__

Solving the first equation for CD and using that in the second equation, we can get an equation for AD, the height of the tower.

CD = AD/tan(67°)

tan(39°)(CD +32) = AD . . . . eliminate fractions in the second equation

tan(39°)(AD/tan(67°) +32) = AD

32·tan(39°) = AD(1 -tan(39°)/tan(67°)) . . . simplify, subtract left-side AD term

32·tan(39°)tan(67°)/(tan(67°) -tan(39°)) = AD . . . . divide by AD coefficient

AD ≈ 39.486 . . . . feet

The tower is about 39.5 feet high.

6 0

2 years ago

Four tomatoes cost $3.40. What is the unit rate?

77julia77 [94]

Answer:

$0.85 per tomato

Step-by-step explanation:

A unit rate is the cost for one item. Like $0.99 for this one bag of skittles.

So $3.40 divided by four is $0.85.

8 0

2 years ago

Read 2 more answers