Question 1

Use technology to solve the following problem: A sample of size 70 will be drawn from a population with mean 25 and standard deviation 11. Find the probability that $\bar { x }$ will be between 23 And 26.

Accepted Answer

The sampling distribution of the sample mean $\bar{x}$ is approximately normal with mean 25 and standard deviation $\frac{11}{\sqrt{70}}$. Using the standard normal distribution, the probability that $\bar{x}$ is between 23 and 26 is approximately 0.7124.

Question 2

A sample of size 50 will be drawn from a population with mean 73 and standard deviation 8. Find the 19th percentile of $\bar { x }$

Accepted Answer

The central limit theorem states that the distribution of sample means is approximately normal, with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of sample size when the sample size is large enough (n ≥ 30). In this case, n = 50, which is large enough.

To find the 19th percentile of the sample means, we need to find the corresponding z-score using the standard normal distribution table.

z-score = invNorm(0.19) = -0.88

Then, we use the formula for the sample mean:

sample mean = population mean + z-score * (population standard deviation / sqrt(sample size))

sample mean = 73 - 0.88 * (8 / sqrt(50))

sample mean ≈ 72.01

Therefore, the 19th percentile of the sample means is approximately 72.01. The closest answer choice is C.

Question 3

The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean $\mu = 40$ and standard deviation $\sigma = 7$ . What proportion of tires have lifetimes of less than 42 Thousand miles?

Accepted Answer

To find the proportion of tires with lifetimes less than 42, we calculate the z-score: $ z = \frac{42 - 40}{7} = \frac{2}{7} \approx 0.2857 $. Using the standard normal distribution table, the cumulative probability for $ z = 0.2857 $ is approximately 0.6141.

Question 4

Use technology to solve the following problem: The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 9.1 pounds and standard deviation 2.4 pounds. A turkey farmer
Wants to provide a money-back guarantee that her 6-week poults will weigh at least a certain Amount. What weight should she guarantee so that she will have to give her customer's money back Only 1% of the time?

Accepted Answer

To find the weight that corresponds to the top 99% of the distribution, we need to find the 1st percentile of a normal distribution with a mean of 9.1 and a standard deviation of 2.4. Using a z-table or calculator, the z-score for the 1st percentile is approximately -2.33. Using the formula: weight = mean + (z-score * standard deviation), we get 9.1 + (-2.33 * 2.4) ≈ 3.51 lb.

Question 5

The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean $\mu = 43$ and standard deviation $\sigma = 5$ . What proportion of tires have lifetimes between 40 and 50 thousand miles?

Accepted Answer

To find the proportion of tires with lifetimes between 40 and 50 thousand miles, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$. For X = 40, $Z = \frac{40 - 43}{5} = -0.6$, and for X = 50, $Z = \frac{50 - 43}{5} = 1.4$. Looking these Z-scores up in the standard normal distribution table gives us the probabilities of 0.7257 and 0.0803, respectively. Subtracting these gives us the proportion of tires in this range: 0.7257 - 0.0803 = 0.6454, which is approximately 0.6450.

Question 6

A sample of size 85 will be drawn from a population with mean 20 and standard deviation 8. Find the probability that $\bar { x }$ will be between 19 and 22.

Accepted Answer

We use the central limit theorem to approximate the sampling distribution of the sample mean as normal with mean = population mean = 20 and standard deviation = population standard deviation / (sample size)^(1/2) = 8 / (85)^(1/2) ≈ 0.867. Then, we standardize the values of the sample mean:
z = (x - mean) / standard deviation = (19 - 20) / 0.867 ≈ -1.15 and z = (22 - 20) / 0.867 ≈ 2.31. Finally, we use the standard normal distribution table or calculator to find the probabilities:
P(-1.15 < z < 2.31) ≈ 0.8642
Therefore, the answer is B.

Question 7

According to a recent study, the weight of male babies less than two months old in the United States is normally distributed with mean 11.4 pounds and standard deviation 2.8 pounds. What proportion Of babies weigh between 10.3 and 12.5 pounds?

Accepted Answer

To find the proportion of babies with weight between 10.3 and 12.5 pounds, we need to find the area under the normal curve between these two values. We can do this by standardizing the values and using a standard normal table or calculator.

First, we calculate the z-scores for the two weights:

z1 = (10.3 - 11.4)/2.8 = -0.39
z2 = (12.5 - 11.4)/2.8 = 0.39

Next, we look up the area to the left of each z-score using a standard normal table or calculator:

P(Z < -0.39) = 0.349
P(Z < 0.39) = 0.651

Finally, we subtract the area to the left of the lower z-score from the area to the left of the higher z-score to get the proportion between the two values:

P(-0.39 < Z < 0.39) = 0.651 - 0.349 = 0.302

Therefore, the proportion of babies weighing between 10.3 and 12.5 pounds is 0.302 or approximately 0.3035. The best answer choice is B.

Question 8

Use technology to solve the following problem: According to a recent study, the weight of male babies less than two months old in the United States is normally distributed with mean 12.0 pounds And standard deviation 2.6 pounds. What proportion of babies weigh less than 10.3 pounds?

Accepted Answer

The answer of Use technology to solve the following problem:...

Question 9

Use technology to solve the following problem: According to a recent study, the weight of male babies less than two months old in the United States is normally distributed with mean 11.4 pounds And standard deviation 2.7 pounds. What proportion of babies weigh between 9.7 and 11.5 pounds?

Accepted Answer

The answer of Use technology to solve the following problem:...

Question 10

Use technology to solve the following problem: The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean $\mu = 40$ and standard deviation $\sigma = 6$ . What Proportion of tires have lifetimes greater than 32 thousand miles?

Accepted Answer

To solve the problem using technology, we can use a standard normal distribution table or calculator. First, we need to standardize the value of 32 using the formula:

z = (x - mu) / sigma

where x is the value we are interested in (32), mu is the mean (40), and sigma is the standard deviation (6). Thus:

z = (32 - 40) / 6 = -1.33

Next, we look up the proportion of the standard normal distribution that is greater than -1.33. Using a table or calculator, we find this to be approximately 0.9088. Therefore, the proportion of tires with lifetimes greater than 32 thousand miles is approximately 0.9088 or 90.88%. The closest choice to this answer is B, with 0.9088 being the correct value rounded to four decimal places.

Question 11

According to a recent study, the weight of male babies less than two months old in the United States is normally distributed with mean 12.0 pounds and standard deviation 2.4 pounds. What proportion Of babies weigh less than 9.1 pounds?

Accepted Answer

The answer of According to a recent study, the weight...

Question 12

According to a recent study, the weight of male babies less than two months old in the United States is normally distributed with mean 11.9 pounds and standard deviation 2.6 pounds. What proportion Of babies weigh more than 10.1 pounds?

Accepted Answer

To find the proportion of babies weighing more than 10.1 pounds, we need to calculate the area under the normal curve to the right of 10.1. To do this, we can standardize the value using the formula:

z = (x - μ) / σ

where x is the given value (10.1), μ is the mean (11.9), and σ is the standard deviation (2.6). Thus,

z = (10.1 - 11.9) / 2.6 = -0.6923

Using a normal distribution table, we can find the area to the right of this z-score, which is equivalent to the proportion of babies weighing more than 10.1 pounds. The table value is 0.7549, which means that approximately 75.49% of male babies less than two months old in the United States weigh more than 10.1 pounds.

Question 13

The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 9.0 pounds and standard deviation 2.4 pounds. A turkey farmer wants to provide a money-back Guarantee that her 6-week poults will weigh at least a certain amount. What weight should she Guarantee so that she will have to give her customer's money back only 1% of the time?

Accepted Answer

To find the weight that corresponds to the top 99% of the distribution, we need to find the z-score that corresponds to the 1st percentile (since 1% of the time the weight will be less than this value). The z-score for the 1st percentile is approximately -2.33. Using the formula: weight = mean + (z-score * standard deviation), we get: weight = 9.0 + (-2.33 * 2.4) = 3.41 lb.

Question 14

The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean $\mu = 39$ and standard deviation $\sigma = 6$ . What proportion of tires have lifetimes greater than 48 Thousand miles?

Accepted Answer

We need to find $P(X > 48)$, where $X$ is a normally distributed random variable with $\mu = 39$ and $\sigma = 6$.

Using z-score, we can standardize the value of 48:
$$z = \frac{48 - 39}{6} = 1.5$$

Now we look up the probability of z being less than 1.5 in the standard normal distribution table which gives us 0.9332.

Since we want the probability that $X$ is greater than 48, we subtract this value from 1:
$$P(X > 48) = 1 - 0.9332 = 0.0668$$

Therefore, the answer is B.

Question 15

Use technology to solve the following problem: The weights of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.8 pounds and standard deviation 1.8 pounds. Find the first Quartile of the weight.

Accepted Answer

The first quartile (Q1) of a normal distribution can be found by using the Z-score table or a calculator with the inverse normal function. The first quartile corresponds to the 25th percentile. Using the formula $Q1 = \mu + Z(\frac{1}{4}) \times \sigma$, where $\mu = 8.8$ (mean) and $\sigma = 1.8$ (standard deviation), and knowing that the Z-score for the 25th percentile is approximately -0.674 (from Z-tables or calculator), we calculate $Q1 = 8.8 + (-0.674 \times 1.8) = 8.8 - 1.2132 = 7.59$ pounds.

Question 16

Use technology to solve the following problem: The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean $\mu = 44$ and standard deviation $\sigma = 4$ . What Proportion of tires have lifetimes of less than 49 thousand miles?

Accepted Answer

To find the proportion of tires with lifetimes less than 49 thousand miles, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X = 49$, $\mu = 44$, and $\sigma = 4$. Plugging in the values, we get $Z = \frac{49 - 44}{4} = 1.25$. Looking up 1.25 in the Z-table or using a normal distribution calculator gives us a proportion of approximately 0.8944, indicating that about 89.44% of the tires have lifetimes of less than 49 thousand miles.

Question 17

Use technology to solve the following problem: According to a recent study, the weight of male babies less than two months old in the United States is normally distributed with mean 11.0 pounds And standard deviation 2.4 pounds. What proportion of babies weigh more than 11.7 pounds?

Accepted Answer

The proportion of babies weighing more than 11.7 pounds is the area under the normal distribution curve to the right of 11.7. Using technology, we can find that this area is 0.3853.

Question 18

A sample of size 39 will be drawn from a population with mean 15 and standard deviation 6. Find the probability that $\bar { x }$ will be greater than 16.

Accepted Answer

The answer of A sample of size 39 will be...

Question 19

Use technology to solve the following problem: The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean $\mu = 37$ and standard deviation $\sigma = 5$ . What Proportion of tires have lifetimes between 33 and 48 thousand miles?

Accepted Answer

We can solve this problem using normal distribution and Z-score. 
First, we need to standardize the values of 33 and 48 using the formula: 
Z = (X - μ) / σ 
For X = 33, Z = (33 - 37) / 5 = -0.8 
For X = 48, Z = (48 - 37) / 5 = 2.2 
Then, we can look up the area under the standard normal curve between -0.8 and 2.2 using a Z-table or a calculator. 
Using a calculator, we get: 
P(-0.8 < Z < 2.2) = 0.7742 
Therefore, the proportion of tires with lifetimes between 33 and 48 thousand miles is 0.7742 or approximately 77.42%. The answer is B.

Question 20

Use technology to solve the following problem: A sample of size 45 will be drawn from a population with mean 94 and standard deviation 14. Find the 32nd percentile of $\bar { x }$

Accepted Answer

The 32nd percentile of the sample mean $\bar{x}$ can be found using the normal distribution. The standard deviation of the sample mean is $\sigma/\sqrt{n} = 14/\sqrt{45}$. Using a standard normal distribution table or calculator, find the z-score for the 32nd percentile, then calculate the corresponding $\bar{x}$ value.

Quiz 7: The Normal Distribution

Use technology to solve the following problem: A sample of size 70 will be drawn from a population with mean 25 and standard deviation 11. Find the probability that xˉ\bar { x }xˉ will be between 23 And 26.

A sample of size 50 will be drawn from a population with mean 73 and standard deviation 8. Find the 19th percentile of xˉ\bar { x }xˉ

The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean μ=40\mu = 40μ=40 and standard deviation σ=7\sigma = 7σ=7 . What proportion of tires have lifetimes of less than 42 Thousand miles?

The lifetime of a certain type of automobile tire (in thousands of miles) is normally distributed with mean μ=43\mu = 43μ=43 and standard deviation σ=5\sigma = 5σ=5 . What proportion of tires have lifetimes between 40 and 50 thousand miles?

A sample of size 85 will be drawn from a population with mean 20 and standard deviation 8. Find the probability that xˉ\bar { x }xˉ will be between 19 and 22.