CALCU-Solve question 1,2,3,4Skill-theet[7]Exercise

Solve question 1,2,3,4

Solution by Steps

step 1

Define the total time taken for a meal to be served as the sum of two independent normally distributed variables: preparation time and cooking time

step 2

Calculate the mean of the total time by adding the means of the preparation time and cooking time:

12 + 14 = 26

minutes

step 3

Calculate the variance of the total time by adding the variances of the preparation time and cooking time:

3^2 + 3^2 = 18

step 4

Calculate the standard deviation of the total time by taking the square root of the variance:

\sqrt{18} = 4.2426

minutes

step 5

Convert the problem to a standard normal distribution problem by finding the Z-score for 30 minutes:

Z = \frac{30 - 26}{4.2426} = 0.9428

step 6

Look up the Z-score in the standard normal distribution table or use a calculator to find the probability of a Z-score greater than 0.9428

step 7

Subtract this probability from 1 to find the probability that a diner will have to wait more than 30 minutes

[question 1] Answer

The probability that a diner will have to wait more than 30 minutes for their meal to be served is approximately 0.1729.

Key Concept

Sum of Independent Normal Variables

Explanation

When adding two independent normal variables, the means add and the variances add. The resulting distribution is also normal with the new mean and standard deviation.

Solution by Steps

step 1

Define the total voltage as the sum of two independent normally distributed variables: voltage of type A and type B batteries

step 2

Calculate the mean of the total voltage by adding the means of type A and type B batteries:

5.0 + 8.0 = 13.0

volts

step 3

Calculate the variance of the total voltage by adding the variances of type A and type B batteries:

0.0225 + 0.04 = 0.0625

step 4

Calculate the standard deviation of the total voltage by taking the square root of the variance:

\sqrt{0.0625} = 0.25

volts

step 5

Convert the problem to a standard normal distribution problem by finding the Z-score for 13.4 volts:

Z = \frac{13.4 - 13.0}{0.25} = 1.6

step 6

Look up the Z-score in the standard normal distribution table or use a calculator to find the probability of a Z-score greater than 1.6

step 7

Subtract this probability from 1 to find the probability that the combined voltage exceeds 13.4 volts

[question 2] Answer

The probability that the combined voltage exceeds 13.4 volts is approximately 0.0548.

Key Concept

Sum of Independent Normal Variables

Explanation

The combined voltage of two independent batteries is also normally distributed, with the mean and standard deviation calculated by summing the individual means and variances, respectively.

Solution by Steps

step 1

Define the difference in scores as a normally distributed variable since the scores are independent and normally distributed

step 2

Calculate the mean of the difference by subtracting the English mean from the mathematics mean:

63 - 68 = -5

step 3

Calculate the variance of the difference by adding the variances of the two scores:

10^2 + 7^2 = 149

step 4

Calculate the standard deviation of the difference by taking the square root of the variance:

\sqrt{149} = 12.2066

step 5

Convert the problem to a standard normal distribution problem by finding the Z-score for a difference of 0:

Z = \frac{0 - (-5)}{12.2066} = 0.4095

step 6

Look up the Z-score in the standard normal distribution table or use a calculator to find the probability of a Z-score greater than 0.4095

step 7

This probability is the same as the probability that a student's mathematics score is higher than their English score

[question 3] Answer

The probability that a student's mathematics score is higher than their English score is approximately 0.3413.

Key Concept

Difference of Independent Normal Variables

Explanation

When subtracting one normal variable from another, the means subtract and the variances add. The resulting distribution is also normal with the new mean and standard deviation.

Solution by Steps

step 1

Define the clearance as the difference between the inner diameter of component B and the outer diameter of component A, which are independent normally distributed variables

step 2

Calculate the mean of the clearance by subtracting the mean of component A from the mean of component B:

0.428 - 0.425 = 0.003

cm

step 3

Calculate the variance of the clearance by adding the variances of component A and component B:

0.0001 + 0.0004 = 0.0005

step 4

Calculate the standard deviation of the clearance by taking the square root of the variance:

\sqrt{0.0005} = 0.0224

cm

step 5

Convert the problem to a standard normal distribution problem by finding the Z-score for a clearance of 0:

Z = \frac{0 - 0.003}{0.0224} = -0.1339

step 6

Look up the Z-score in the standard normal distribution table or use a calculator to find the probability of a Z-score less than -0.1339

step 7

This probability is the same as the probability that component A will not fit inside component B

[question 4] Answer

The probability that component A will not fit inside component B is approximately 0.4468.

Key Concept

Difference of Independent Normal Variables

Explanation

The clearance between two components is normally distributed, with the mean and standard deviation calculated by subtracting the means and adding the variances of the individual components' dimensions.

solve question 5，6,7,8

Solution by Steps

step 1

Define the random variable

X

as the difference in scores between the two students

step 2

Since both scores are normally distributed with the same mean

\mu = 25000

and standard deviation

\sigma = 3000

, the difference

X

is also normally distributed with mean

\mu_X = 0

and standard deviation

\sigma_X = \sqrt{2} \cdot 3000

step 3

Calculate the z-score for a difference of 7500 points:

z = \frac{7500}{\sigma_X}

step 4

Use the standard normal distribution table to find the probability that |X| > 7500

[question 5] Answer

The probability that the students' scores will differ by more than 7500 points is the sum of the probabilities that X > 7500 and X < -7500.

Key Concept

Difference of two independent normal random variables

Explanation

When two independent normal random variables with the same mean and standard deviation are subtracted, the resulting variable is also normally distributed with a mean of zero and a standard deviation that is the square root of the sum of the variances of the two variables.

Question 6:

step 1

Define the random variable

Y

as the total weight of a bag of six bananas

step 2

Since the weight of one banana is normally distributed with mean

\mu = 180

g and standard deviation

\sigma = 20

g, the sum of six bananas will have mean

\mu_Y = 6 \cdot 180

g and standard deviation

\sigma_Y = \sqrt{6} \cdot 20

g

step 3

Convert the weight limit of 1 kg to grams:

1000

g

step 4

Calculate the z-score for the weight limit:

z = \frac{1000 - \mu_Y}{\sigma_Y}

step 5

Use the standard normal distribution table to find the probability that Y < 1000 g

[question 6] Answer

The probability that the weight of a randomly chosen bag of six bananas is less than 1 kg is found using the z-score and the standard normal distribution table.

Key Concept

Sum of independent normal random variables

Explanation

The sum of independent normal random variables is also normally distributed, with the mean being the sum of the means and the variance being the sum of the variances.

Question 7:

step 1

Define the random variable

Z

as the total weight of the people in the elevator

step 2

Since the weight of one person is normally distributed with mean

\mu = 82

kg and standard deviation

\sigma = 9

kg, the sum of

n

people will have mean

\mu_Z = n \cdot 82

kg and standard deviation

\sigma_Z = \sqrt{n} \cdot 9

kg

step 3

Calculate the z-score for the weight limit of the elevator:

z = \frac{680 - \mu_Z}{\sigma_Z}

step 4

Use the standard normal distribution table to find the value of

n

that makes us 99% sure the elevator does not exceed capacity

[question 7] Answer

The maximum number of people who can get into the elevator to be at least 99% sure that the elevator does not exceed capacity is determined by solving for

n

using the z-score and the standard normal distribution table.

Key Concept

Normal approximation for the sum of weights

Explanation

The total weight of a group of people can be approximated by a normal distribution, and we can use the z-score to find the probability that the total weight is below a certain threshold.

Question 8:

step 1

Define the random variable

W

as the total life of the alarm system with 20 batteries

step 2

Since each battery has a mean life of 7 hours and standard deviation of 0.5 hours, and they operate independently, the sum of 20 batteries will have mean

\mu_W = 20 \cdot 7

hours and standard deviation

\sigma_W = \sqrt{20} \cdot 0.5

hours

step 3

Calculate the z-score for the system life of 145 hours:

z = \frac{145 - \mu_W}{\sigma_W}

step 4

Use the standard normal distribution table to find the probability that W > 145 hours

[question 8] Answer

The probability that the alarm system is still working after 145 hours is found using the z-score and the standard normal distribution table.

Key Concept

Sum of independent normal random variables for battery life

Explanation

The total life of the alarm system can be modeled as the sum of the lives of the individual batteries, which is a normally distributed random variable with its own mean and standard deviation.

Solve questions 5,6,7,8

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