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Blizzard [7]
3 years ago
13

Chapter 7, Section 7.2, Question P9 The process for manufacturing a ball bearing results in weights that have an approximately n

ormal distribution with mean 0.15 g and standard deviation 0.005 g. a. If you select one ball bearing at random, what is the probability that it weighs less than 0.149 g? (round to four decimal places) b. If you select four ball bearings at random, what is the probability that their mean weight is less than 0.149 g? (round to four decimal places) c. If you select ten ball bearings at random, what is the probability that their mean weight is less than 0.149 g? (round to four decimal places)
Mathematics
1 answer:
bogdanovich [222]3 years ago
8 0

a. Let X be a random variable representing the weight of a ball bearing selected at random. We're told that X\sim\mathcal N(0.15,0.005^2), so

\mathrm P(X

where Z\sim\mathcal N(0,1). This probability is approximately

\mathrm P(Z

b. Let X_i be a random variable representing the weight of the i-th ball that is selected, and let Y be the mean of these 4 weights,

Y=\dfrac{X_1+X_2+X_3+X_4}4

The sum of normally distributed random variables is a random variable that also follows a normal distribution,

X_1+X_2+X_3+X_4\sim\mathcal N(4\cdot0.15,4\cdot0.005^2)

so that

Y\sim\mathcal N(0.15,0.005^2)

Then

\mathrm P(Y

c. Same as (b).

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Answer:

12 servings

Step-by-step explanation:

Luke's turkey chili recipe calls for 2.5 pounds of ground turkey for every 5 servings. How many servings can he make if he has 6 pounds of ground turkey?

From above question:

2.5 pounds of ground turkey = 5 servings

6 pounds of ground turkey = x

Cross Multiply

2.5 × x = 6 × 5

x = 6 × 5/2.5

x = 12 servings

Hence, 6 pounds of ground turkey can make 12 servings.

Completing the explanation.

The equation is

5x = 2.5y

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x = represent how many pounds of ground turkey he has = 6 pounds

y = to represent how many servings he can make

Hence,

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y = 5 × 6/2.5

y = 12 servings

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Answer:

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U 2 can help me by marking as brainliest.........

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