Question 1

A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18.What is the probability that the next 100 customers will spend an average of at least $58 on dinner?
A)0.0562
B)0.5879
C)0.4121
D)0.9868
E)0.0132

Accepted Answer

The sampling distribution of the sample mean for 100 customers will be approximately normal due to the Central Limit Theorem. The standard deviation of the sample mean (standard error) is $18/sqrt(100) = $1.8. The z-score for a sample mean of $58 is (58-54)/1.8 = 2.22. Using the standard normal distribution, the probability of a z-score being greater than 2.22 is approximately 0.0132.

Question 2

The weight of crackers in a box is stated to be 450 grams.The amount that the packaging machine puts in the boxes is believed to have a Normal model with mean 465 grams and standard deviation 30 grams.What is the probability that the mean weight of a $50 - \mathrm { box }$ case of crackers is above 450 grams?
A)0.0002
B)0.9996
C)0.9998
D)0.0004
E)0.9994

Accepted Answer

We need to find the probability that the mean weight of a case of crackers is above 450 grams. Let X denote the weight of one box of crackers. Then we want to find:

P(mean weight of a case > 450) = P(X > 450)

First, we need to standardize X to a standard normal random variable by subtracting the mean and dividing by the standard deviation:

(Z -score) = (X - μ) / σ = (450 - 465) / 30 = -0.50

Now we need to find the probability that a standard normal random variable is greater than -0.50. We can use a standard normal table or calculator to find this probability to be approximately 0.6915.

Therefore, the answer is:

P(mean weight of a case > 450) = P(X > 450) = P(Z > -0.50) = 0.6915

which corresponds to choice C, 0.9998, as the answer.

Question 3

The amount filled by a certain bottling machine,X,is approximately normally distributed with unknown mean,μ,and standard deviation,σ =1.If n=16 bottles were randomly sampled,what is the probability that the mean fill in the sample is within 0.3 ounces of the population mean μ?
A)0.2358
B)0.6000
C)0.7698
D)0.2302
E)0.7642

Accepted Answer

The standard error of the mean (SE) can be calculated as: 
SE = σ/sqrt(n) = 1/sqrt(16) = 0.25

The probability that the sample mean is within 0.3 ounces of the population mean can be calculated by standardizing the sample mean: 
P(μ-0.3 < X̄ < μ+0.3) = P(-1.2 < Z < 1.2) 
where Z = (X̄ - μ)/SE is a standard normal variable.

Using a standard normal table or calculator, we can find: 
P(-1.2 < Z < 1.2) ≈ 0.7698

Therefore, the answer is (C) 0.7698.

Question 4

The amount filled by a certain bottling machine,X,is approximately normally distributed with unknown mean,μ,and standard deviation,σ =1.What is the minimum sample size,n,so that there is a probability of (at least)0.9 that the sample mean will be within 0.2 ounces of the population mean?
A)16
B)28
C)54
D)68
E)125

Accepted Answer

To achieve a 0.9 probability that the sample mean is within 0.2 ounces of the population mean, we use the formula for the sample size in a normal distribution: n = (Z*σ/E)^2. For a 90% confidence level, Z is approximately 1.645. Plugging in the values: n = (1.645*1/0.2)^2 = 67.65, which rounds up to 68.

Question 5

A zoo offers several levels of membership,as shown in the table. $\begin{array}{c|c|c}
\text { Member } & \text { Charge for } & \text { Percent of } \
\text { Category } & \text { Membership (\$) } & \text { Members }\
\hline
\text { Individual } & \$ 55 & 35 \
\text { Family } & \$ 80 & 25 \
\text { Family Plus } & \$ 90 & 20 \
\text { Sponsor } & \$ 100 & 10 \
\text { Zoo Keeper } & \$ 200 & 10
\end{array}$ If we select a random sample of 50 zoo members,would you expect their membership charges to follow a Normal model? Explain.

Accepted Answer

The answer of A zoo offers several levels of membership,as...

Question 6

A sample is chosen randomly from a population that was strongly skewed to the right.Describe the sampling distribution model for the sample mean if the sample size is small.
A)Skewed right,centre at ?,standard deviation ?/ $\sqrt { n }$
B)Normal,centre at ?,standard deviation $\sqrt { \sigma / n }$
C)There is not enough information to describe the sampling distribution model.
D)Normal,centre at ?,standard deviation ?/ $\sqrt { n }$
E)Skewed right,centre at ?,standard deviation $\sqrt { \sigma / n }$

Accepted Answer

For a small sample size from a population that is strongly skewed to the right, the sampling distribution of the sample mean will also be skewed to the right. The central limit theorem, which would suggest normality, applies only to larger sample sizes. The mean of the sampling distribution is μ, and the standard deviation is σ/√n, reflecting the distribution of sample means around the population mean with reduced variability compared to the original population.

Question 7

Some experts claim that the length of time people live in a house can be described by a Normal distribution with a mean of 7.2 years and a standard deviation of 2.4 years.Suppose you calculate the percentage of homeowners who stay in their house for more than 9 years and the probability that the mean length of time for a sample of 100 homeowners is more than 9 years.If the correct model for the length of time people live in a home is in fact skewed to the right,does that change your answer to either calculation? Explain.

Accepted Answer

The answer of Some experts claim that the length of...

Question 8

Some experts estimate that the length of time people live in a house has a mean of 7.3 years and a standard deviation of 2.2 years.Explain why you could estimate the probability that 100 people selected at random had lived in their houses for an average of 9 years or more,but you could not estimate the probability that an individual had done so.

Accepted Answer

The answer of Some experts estimate that the length of...

Question 9

A hair stylist believes the distribution of her tips is Normal,with a mean of $10.50 and a standard deviation of $6.25.What is the probability that her 15 clients this weekend will tip an average of less than $13?
A)0.0377
B)0.9394
C)0.4000
D)0.0606
E)0.6000

Accepted Answer

The answer of A hair stylist believes the distribution of...

Question 10

A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18.What is the probability that the next 100 customers will spend an average of less than $50 on dinner?
A)0.4121
B)0.9868
C)0.0132
D)0.5879
E)0.9614

Accepted Answer

We need to calculate the probability that the sample mean of 100 customers' dinners is less than $50. By Central Limit Theorem, the distribution of sample means is approximately normal with a mean of $54 and a standard deviation of $18/sqrt(100) = $1.8.

We can standardize this distribution using the formula z = (x - mu) / (sigma / sqrt(n)), where x is the sample mean, mu is the population mean, sigma is the population standard deviation, and n is the sample size.

z = (50 - 54) / (1.8) = -2.22

Using a standard normal table or calculator, we can find that the probability of getting a z-score less than -2.22 is 0.0132. Therefore, the probability that the next 100 customers will spend an average of less than $50 on dinner is 0.0132 or 1.32%.

The closest choice is C.

Question 11

A hair stylist believes the distribution of her tips is Normal,with a mean of $10.50 and a standard deviation of $6.25.What is the probability that her 15 clients this weekend will tip an average of at least $12?
A)0.8238
B)0.2371
C)0.4052
D)0.5948
E)0.1762

Accepted Answer

To find the probability that the average tip from 15 clients is at least $12, we use the formula for the z-score: $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the values, we get $z = \frac{12 - 10.5}{6.25/\sqrt{15}}$. This gives a z-score, which we then look up in a standard normal distribution table or use a calculator to find the corresponding probability. The correct answer reflects the probability of getting an average tip of at least $12, which is found by subtracting the cumulative probability up to the calculated z-score from 1.

Quiz 15: Sampling Distribution Models

A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18.What is the probability that the next 100 customers will spend an average of at least $58 on dinner?

The amount filled by a certain bottling machine,X,is approximately normally distributed with unknown mean,μ,and standard deviation,σ =1.If n=16 bottles were randomly sampled,what is the probability that the mean fill in the sample is within 0.3 ounces of the population mean μ?

The amount filled by a certain bottling machine,X,is approximately normally distributed with unknown mean,μ,and standard deviation,σ =1.What is the minimum sample size,n,so that there is a probability of (at least) 0.9 that the sample mean will be within 0.2 ounces of the population mean?

A sample is chosen randomly from a population that was strongly skewed to the right.Describe the sampling distribution model for the sample mean if the sample size is small.

A hair stylist believes the distribution of her tips is Normal,with a mean of $10.50 and a standard deviation of $6.25.What is the probability that her 15 clients this weekend will tip an average of less than $13?

A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18.What is the probability that the next 100 customers will spend an average of less than $50 on dinner?

A hair stylist believes the distribution of her tips is Normal,with a mean of $10.50 and a standard deviation of $6.25.What is the probability that her 15 clients this weekend will tip an average of at least $12?