Question 1

At a cell phone assembly plant, 78% of the cell phone keypads pass inspection. A random sample of 104 keypads is analyzed. Find the standard deviation

Accepted Answer

To find the standard deviation, we use the formula for the standard deviation of a binomial distribution: 
standard deviation = sqrt(p*(1-p)/n), where p is the proportion of successes (in this case, the proportion of keypads that pass inspection, which is 0.78) and n is the sample size (which is 104). 
Plugging in the values, we get: 
standard deviation = sqrt(0.78*(1-0.78)/104) = 0.041 
Therefore, the answer is D.

Question 2

For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 92 diamonds is analyzed. Find the probability that more than 79% of the sample Diamonds fail to qualify as gemstone grade.

Accepted Answer

Let X be the number of diamonds that fail to qualify as gemstone grade in the sample. Then X follows a binomial distribution with n = 92 and p = 0.81.

We want to find P(X > 0.79*92) = P(X > 72.68). Since X is discrete, we need to use a continuity correction. That is, we need to find P(X > 73).

Using the binomial probability formula or a calculator, we get:

P(X > 73) = 1 - P(X ≤ 73) = 1 - binomcdf(92, 0.81, 73) ≈ 0.6879

Therefore, the best choice is A.

Question 3

At a cell phone assembly plant, 84% of the cell phone keypads pass inspection. A random sample of 106 keypads is analyzed. Find the mean $\underset { p } { \mu ^ { \wedge } }$

Accepted Answer

The mean of a binomial distribution is given by
$\mu=p$
where $p$ is the probability of success. 
In this case, the probability of a keypad passing inspection is 0.84. Therefore, the mean is $\mu=0.84$.

Question 4

For a particular diamond mine, 80% of the diamonds fail to qualify as "gemstone grade". A random sample of 94 diamonds is analyzed. Find the probability that the proportion of the sample diamonds That fail to qualify as gemstone grade is between 0.76 and 0.83.

Accepted Answer

This problem can be solved using the Central Limit Theorem (CLT) for proportions. Given that 80% of the diamonds fail to qualify, the population proportion (p) is 0.80, and the standard deviation of the sampling distribution of the sample proportion ($\sigma_{\hat{p}}$) can be calculated using the formula $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$, where n is the sample size (94 in this case). To find the probability that the sample proportion ($\hat{p}$) falls between 0.76 and 0.83, we convert these proportions to z-scores and use the standard normal distribution to find the probability between these z-scores. The calculation results in a probability of 0.6013, indicating that there's about a 60.13% chance the proportion of diamonds that fail to qualify as gemstone grade in the sample is between 0.76 and 0.83.

Question 5

Use technology to solve the following problem: A survey reported that in a recent year, the mean serum cholesterol level in milligrams per deciliter for U.S. adults was 195 with a standard deviation Of 35. A random sample of 110 adults was chosen. Find the 77th percentile of the sample mean.

Accepted Answer

To find the 77th percentile of the sample mean, we need to use the standard normal distribution since the sample size is greater than 30 (central limit theorem).

First, we need to find the standard error of the mean:
standard error of the mean = standard deviation / √sample size
standard error of the mean = 35 / √110
standard error of the mean ≈ 3.34

Next, we need to find the z-score associated with the 77th percentile:
z-score = invNorm(0.77)
z-score ≈ 0.61

Finally, we can use the formula for the standard normal distribution:
z = (x̄ - μ) / (σ/√n)
0.61 = (x̄ - 195) / (35/√110)

Solving for x̄ gives us:
x̄ ≈ 197.5

Therefore, the 77th percentile of the sample mean is approximately 197.5. The best choice is C.

Question 6

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } .$ $n = 108 , p = 0.1 ; P ( 0.06 < \hat { p } < 0.12 )$

Accepted Answer

The answer of Using technology, use the Central Limit Theorem...

Question 7

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p }$ $n = 126 , p = 0.4 ; P ( 0.35 < \hat { p } < 0.42 )$

Accepted Answer

According to the Central Limit Theorem, when the sample size is sufficiently large (n > 30), the distribution of sample proportions is approximately normal with mean equal to the population proportion (p) and standard deviation equal to the square root of [(p * (1-p))/n].

In this case, n = 126 and p = 0.4, so the mean of the sample proportion is 0.4 and the standard deviation is sqrt[(0.4 * 0.6)/126] = 0.0509.

To find P(0.35 < p-hat < 0.42), we need to standardize the values using the formula z = (x - mu) / sigma, where x is the value in question, mu is the mean, and sigma is the standard deviation.

P(0.35 < p-hat < 0.42) can be rewritten as P((0.35 - 0.4) / 0.0509 < z < (0.42 - 0.4) / 0.0509) = P(-0.98 < z < 0.39).

Using a standard normal table or calculator, we can find that P(-0.98 < z < 0.39) is approximately 0.5522.

Therefore, the answer is C.

Question 8

Use technology to solve the following problem: A survey reported that in a recent year, the mean serum cholesterol level in milligrams per deciliter for U.S. adults was 201 with a standard deviation
Of 41. A random sample of 114 adults was chosen. What is the probability that the mean Cholesterol level is between 198 and 203?

Accepted Answer

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

Question 9

Use technology to solve the following problem: For a particular diamond mine, 78% of the diamonds fail to qualify as "gemstone grade". A random sample of 90 diamonds is analyzed. Find The probability that the proportion of the sample diamonds that fail to qualify as gemstone grade is
Between 0.73 and 0.81.

Accepted Answer

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

Question 10

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p }$ $n = 195 , p = 0.48 ; P ( \hat { p } < 0.5 )$

Accepted Answer

The Central Limit Theorem allows us to use the normal distribution to approximate the sampling distribution of the sample proportion. The standard deviation of the sample proportion is $\sqrt{\frac{p(1-p)}{n}}$. Using this, we calculate the z-score for $\hat{p} = 0.5$ and find the probability $P(\hat{p} < 0.5)$ using the standard normal distribution table, which corresponds to approximately 0.7123.

Question 11

At a cell phone assembly plant, 79% of the cell phone keypads pass inspection. A random sample of 103 keypads is analyzed. Find the probability that more than 83% of the sample keypads pass Inspection.

Accepted Answer

The sample size is larger than 30, so we can use the normal distribution to approximate the binomial distribution. 
The mean of the sample proportion is 0.79, and the standard deviation of the sample proportion is:
sqrt(0.79*0.21/103) = 0.046
To find the probability that more than 83% of the sample keypads pass inspection, we need to standardize the value:
z = (0.83 - 0.79)/0.046 = 0.87
Using a standard normal distribution table or calculator, we find that the probability of z being greater than 0.87 is 0.1587. Therefore, the answer is B.

Question 12

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } \text {. }$ $n = 114 , p = 0.51 ; P ( \hat { p } > 0.47 )$

Accepted Answer

We can use the normal distribution approximation to the binomial distribution by applying Central Limit Theorem, since $np=58.14>10$ and $n(1-p)=55.86>10$.
The standardized test statistic is:
$$
\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}=\frac{0.47-0.51}{\sqrt{\frac{0.51(0.49)}{114}}}\approx -1.84
$$
Using normal distribution table, we get the probability as $P(\hat p>0.47)=P\left(Z>\frac{-1.84+0.005}{1}\right)\approx 0.8035$. Therefore, the answer is A).

Question 13

For a particular diamond mine, 82% of the diamonds fail to qualify as "gemstone grade". A random sample of 102 diamonds is analyzed. Find the probability that less than 88% of the sample diamonds Fail to qualify as gemstone grade.

Accepted Answer

The probability of a diamond failing to qualify as gemstone grade is 0.82. The probability of a diamond qualifying as gemstone grade is 1 - 0.82 = 0.18. The probability of less than 88% of the sample diamonds failing to qualify as gemstone grade is the probability of 0, 1, 2, ..., 17 diamonds failing to qualify as gemstone grade. This is given by the binomial distribution:P(X < 18) = ΣP(X = k) for k = 0, 1, 2, ..., 17= Σ(102 choose k) * (0.82)^k * (0.18)^(102-k) for k = 0, 1, 2, ..., 17= 0.9429Therefore, the probability that less than 88% of the sample diamonds fail to qualify as gemstone grade is 0.9429.

Question 14

A survey reported that in a recent year, the mean serum cholesterol level in milligrams per deciliter for U.S. adults was 204 with a standard deviation of 37. A random sample of 116 adults was Chosen. Find the 33rd percentile of the sample mean.

Accepted Answer

Since the sample size is greater than 30, we can use the Central Limit Theorem to find the distribution of the sample mean. The z-score corresponding to the 33rd percentile is -0.44 (found using a standard normal distribution table).

Using the formula for the standard error of the mean:

SE = standard deviation / √n 
SE = 37 / √116 
SE = 3.44

We can now use the formula for the confidence interval of the sample mean:

Sample mean ± (z-score)(SE)

Plugging in the values:

Sample mean ± (-0.44)(3.44) 
204 ± (-1.51) 
202.49

Therefore, the 33rd percentile of the sample mean is approximately 202.5.

Question 15

For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 97 diamonds is analyzed. Find the standard deviation $\begin{array} { c } 
\sigma ^ { \wedge } \
p
\end{array}$

Accepted Answer

Since we are dealing with a binomial distribution (either a diamond qualifies as gemstone grade or it doesn't), the formula for the standard deviation is:
$$
\sigma = \sqrt{np(1-p)}
$$
where n is the sample size and p is the probability of success (qualifying as gemstone grade) in a single trial.

In this case, n = 97 and p = 0.19 (since 81% of diamonds fail to qualify, then 19% qualify), so plugging in these values:
$$
\sigma = \sqrt{(97)(0.19)(0.81)} \approx 0.04
$$

Therefore, the best choice is C.

Question 16

Use technology to solve the following problem: For a particular diamond mine, 80% of the diamonds fail to qualify as "gemstone grade". A random sample of 131 diamonds is analyzed. Find The probability that less than 75% of the sample diamonds fail to qualify as gemstone grade.

Accepted Answer

We can use the binomial distribution to solve this problem. Let X be the number of diamonds in the sample that fail to qualify as gemstone grade. Then X has a binomial distribution with n = 131 and p = 0.8. We want to find P(X < 0.75(131)) = P(X < 98.25).

Using normal approximation to binomial, we can approximate X with a normal distribution with mean np = 131(0.8) = 104.8 and standard deviation sqrt(np(1-p)) = sqrt(131(0.8)(0.2)) = 4.29. Then,

z = (98.25 - 104.8)/4.29 = -1.51

Using a standard normal table or calculator, we find that P(Z < -1.51) = 0.0658. Therefore, the probability that less than 75% of the sample diamonds fail to qualify as gemstone grade is approximately 0.0763.

Hence, the answer is C.

Question 17

For a particular diamond mine, 78% of the diamonds fail to qualify as "gemstone grade". A random sample of 106 diamonds is analyzed. Find the mean $\underset { p } { \mu ^ { \wedge } }$

Accepted Answer

The mean of a binomial distribution is given by $\mu = np$, where $n$ is the number of trials (in this case, 106 diamonds) and $p$ is the probability of success (in this case, the probability of a diamond qualifying as gemstone grade, which is 1 - 0.78 = 0.22). However, the question seems to ask for the mean proportion ($\mu^\wedge$) of diamonds that fail to qualify, which directly corresponds to the given failure rate of 78% or 0.78, not requiring the calculation involving the sample size.

Question 18

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } \text {. }$ $n = 166 , p = 0.61 ; P ( \hat { p } > 0.58 )$

Accepted Answer

The answer of Use the Central Limit Theorem to find...

Question 19

Use technology to solve the following problem: For a particular diamond mine, 82% of the diamonds fail to qualify as "gemstone grade". A random sample of 117 diamonds is analyzed. Find The probability that more than 79% of the sample diamonds fail to qualify as gemstone grade.

Accepted Answer

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

Question 20

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } \text {. }$
 $n = 124 , p = 0.57 ; P ( \hat { p } < 0.63 )$

Accepted Answer

The answer of Using technology, use the Central Limit Theorem...

Quiz 7: The Normal Distribution

At a cell phone assembly plant, 78% of the cell phone keypads pass inspection. A random sample of 104 keypads is analyzed. Find the standard deviation

For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 92 diamonds is analyzed. Find the probability that more than 79% of the sample Diamonds fail to qualify as gemstone grade.

At a cell phone assembly plant, 84% of the cell phone keypads pass inspection. A random sample of 106 keypads is analyzed. Find the mean $\underset { p } { \mu ^ { \wedge } }$

For a particular diamond mine, 80% of the diamonds fail to qualify as "gemstone grade". A random sample of 94 diamonds is analyzed. Find the probability that the proportion of the sample diamonds That fail to qualify as gemstone grade is between 0.76 and 0.83.

Use technology to solve the following problem: A survey reported that in a recent year, the mean serum cholesterol level in milligrams per deciliter for U.S. adults was 195 with a standard deviation Of 35. A random sample of 110 adults was chosen. Find the 77th percentile of the sample mean.

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } .$ $n = 108 , p = 0.1 ; P ( 0.06 < \hat { p } < 0.12 )$

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p }$ $n = 126 , p = 0.4 ; P ( 0.35 < \hat { p } < 0.42 )$

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p }$ $n = 195 , p = 0.48 ; P ( \hat { p } < 0.5 )$

At a cell phone assembly plant, 79% of the cell phone keypads pass inspection. A random sample of 103 keypads is analyzed. Find the probability that more than 83% of the sample keypads pass Inspection.

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } \text {. }$ $n = 114 , p = 0.51 ; P ( \hat { p } > 0.47 )$

For a particular diamond mine, 82% of the diamonds fail to qualify as "gemstone grade". A random sample of 102 diamonds is analyzed. Find the probability that less than 88% of the sample diamonds Fail to qualify as gemstone grade.

A survey reported that in a recent year, the mean serum cholesterol level in milligrams per deciliter for U.S. adults was 204 with a standard deviation of 37. A random sample of 116 adults was Chosen. Find the 33rd percentile of the sample mean.

For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 97 diamonds is analyzed. Find the standard deviation $\begin{array} { c } \sigma ^ { \wedge } \\p\end{array}$

Use technology to solve the following problem: For a particular diamond mine, 80% of the diamonds fail to qualify as "gemstone grade". A random sample of 131 diamonds is analyzed. Find The probability that less than 75% of the sample diamonds fail to qualify as gemstone grade.

For a particular diamond mine, 78% of the diamonds fail to qualify as "gemstone grade". A random sample of 106 diamonds is analyzed. Find the mean $\underset { p } { \mu ^ { \wedge } }$

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } \text {. }$ $n = 166 , p = 0.61 ; P ( \hat { p } > 0.58 )$

Use technology to solve the following problem: For a particular diamond mine, 82% of the diamonds fail to qualify as "gemstone grade". A random sample of 117 diamonds is analyzed. Find The probability that more than 79% of the sample diamonds fail to qualify as gemstone grade.

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } \text {. }$ $n = 124 , p = 0.57 ; P ( \hat { p } < 0.63 )$

Quiz 7: The Normal Distribution

At a cell phone assembly plant, 78% of the cell phone keypads pass inspection. A random sample of 104 keypads is analyzed. Find the standard deviation

For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 92 diamonds is analyzed. Find the probability that more than 79% of the sample Diamonds fail to qualify as gemstone grade.

At a cell phone assembly plant, 84% of the cell phone keypads pass inspection. A random sample of 106 keypads is analyzed. Find the mean μ∧p\underset { p } { \mu ^ { \wedge } }pμ∧​

For a particular diamond mine, 80% of the diamonds fail to qualify as "gemstone grade". A random sample of 94 diamonds is analyzed. Find the probability that the proportion of the sample diamonds That fail to qualify as gemstone grade is between 0.76 and 0.83.

Use technology to solve the following problem: A survey reported that in a recent year, the mean serum cholesterol level in milligrams per deciliter for U.S. adults was 195 with a standard deviation Of 35. A random sample of 110 adults was chosen. Find the 77th percentile of the sample mean.

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is p^.\hat { p } .p^​. n=108,p=0.1;P(0.06<p^<0.12) n = 108 , p = 0.1 ; P ( 0.06 < \hat { p } < 0.12 ) n=108,p=0.1;P(0.06<p^​<0.12)

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is p^\hat { p }p^​ n=126,p=0.4;P(0.35<p^<0.42) n = 126 , p = 0.4 ; P ( 0.35 < \hat { p } < 0.42 ) n=126,p=0.4;P(0.35<p^​<0.42)

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is p^\hat { p }p^​ n=195,p=0.48;P(p^<0.5) n = 195 , p = 0.48 ; P ( \hat { p } < 0.5 ) n=195,p=0.48;P(p^​<0.5)

At a cell phone assembly plant, 79% of the cell phone keypads pass inspection. A random sample of 103 keypads is analyzed. Find the probability that more than 83% of the sample keypads pass Inspection.

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is p^. \hat { p } \text {. }p^​. n=114,p=0.51;P(p^>0.47) n = 114 , p = 0.51 ; P ( \hat { p } > 0.47 ) n=114,p=0.51;P(p^​>0.47)

For a particular diamond mine, 82% of the diamonds fail to qualify as "gemstone grade". A random sample of 102 diamonds is analyzed. Find the probability that less than 88% of the sample diamonds Fail to qualify as gemstone grade.

A survey reported that in a recent year, the mean serum cholesterol level in milligrams per deciliter for U.S. adults was 204 with a standard deviation of 37. A random sample of 116 adults was Chosen. Find the 33rd percentile of the sample mean.

For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 97 diamonds is analyzed. Find the standard deviation σ∧p\begin{array} { c } \sigma ^ { \wedge } \\p\end{array}σ∧p​

Use technology to solve the following problem: For a particular diamond mine, 80% of the diamonds fail to qualify as "gemstone grade". A random sample of 131 diamonds is analyzed. Find The probability that less than 75% of the sample diamonds fail to qualify as gemstone grade.

For a particular diamond mine, 78% of the diamonds fail to qualify as "gemstone grade". A random sample of 106 diamonds is analyzed. Find the mean μ∧p\underset { p } { \mu ^ { \wedge } }pμ∧​

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is p^. \hat { p } \text {. }p^​. n=166,p=0.61;P(p^>0.58) n = 166 , p = 0.61 ; P ( \hat { p } > 0.58 ) n=166,p=0.61;P(p^​>0.58)

Use technology to solve the following problem: For a particular diamond mine, 82% of the diamonds fail to qualify as "gemstone grade". A random sample of 117 diamonds is analyzed. Find The probability that more than 79% of the sample diamonds fail to qualify as gemstone grade.

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is p^. \hat { p } \text {. }p^​. n=124,p=0.57;P(p^<0.63) n = 124 , p = 0.57 ; P ( \hat { p } < 0.63 ) n=124,p=0.57;P(p^​<0.63)

At a cell phone assembly plant, 84% of the cell phone keypads pass inspection. A random sample of 106 keypads is analyzed. Find the mean $\underset { p } { \mu ^ { \wedge } }$

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } .$ $n = 108 , p = 0.1 ; P ( 0.06 < \hat { p } < 0.12 )$

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p }$ $n = 126 , p = 0.4 ; P ( 0.35 < \hat { p } < 0.42 )$

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p }$ $n = 195 , p = 0.48 ; P ( \hat { p } < 0.5 )$

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } \text {. }$ $n = 114 , p = 0.51 ; P ( \hat { p } > 0.47 )$

For a particular diamond mine, 81% of the diamonds fail to qualify as "gemstone grade". A random sample of 97 diamonds is analyzed. Find the standard deviation $\begin{array} { c } \sigma ^ { \wedge } \\p\end{array}$

For a particular diamond mine, 78% of the diamonds fail to qualify as "gemstone grade". A random sample of 106 diamonds is analyzed. Find the mean $\underset { p } { \mu ^ { \wedge } }$

Use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } \text {. }$ $n = 166 , p = 0.61 ; P ( \hat { p } > 0.58 )$

Using technology, use the Central Limit Theorem to find the indicated probability. The sample size is n, the population proportion is p, and the sample proportion is $\hat { p } \text {. }$ $n = 124 , p = 0.57 ; P ( \hat { p } < 0.63 )$