MATH 1680 Exam 3

The graph of the normal curve slides right.

What happens to the graph of the normal curve as the mean​ increases? A. The graph of the normal curve compresses and becomes steeper. B. The graph of the normal curve slides right. C. The graph of the normal curve flattens out and becomes wider. D. The graph of the normal curve slides left. E. Nothing happens to the graph of the normal curve.

The graph of the normal curve compresses and becomes steeper.

What happens to the graph of the normal curve as the standard deviation​ decreases? A. The graph of the normal curve slides left. B. The graph of the normal curve flattens out and becomes wider. C. The graph of the normal curve slides right. D. The graph of the normal curve compresses and becomes steeper. E. Nothing happens to the graph of the normal curve.

No​, because the histogram does not have the shape of a normal curve.

The relative frequency histogram represents the length of phone calls on​ George's cell phone during the month of September. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable. A. ​No, because the histogram has the shape of a normal curve. B. ​Yes, because the histogram does not have the shape of a normal curve. C. No​, because the histogram does not have the shape of a normal curve. D. Yes​, because the histogram has the shape of a normal curve. E. Yes​, because the histogram is not symmetric about its mean.

The histogram is not bell dash shaped comma so a normal distribution could not be used as a model for the variable.

A study was conducted that resulted in the following relative frequency histogram. Determine whether or not the histogram indicates that a normal distribution could be used as a model for the variable. A.The histogram is not​ bell-shaped, so a normal distribution could be used as a model for the variable. B. The histogram is bell dash shaped comma so a normal distribution could be used as a model for the variable. C. The histogram is​ bell-shaped, so a normal distribution could not be used as a model for the variable. D. The histogram is not bell dash shaped comma so a normal distribution could not be used as a model for the variable.

15.87% of cell phone plans cost less than ​$49 per month.

Suppose the monthly charge for cell phone plans is normally distributed with mean μ = $ 66 and standard deviation σ = $17. ​ Suppose the area under the normal curve to the left of X = $49 is 0.1587. Provide an interpretation of this result. A. ​15.87% of cell phone plans cost less than ​$49 per month. B. The probability is 0.1587 that a randomly selected cell phone plan is more than $ 49 per month. C. ​15.87% of cell phone plans cost more than ​$49 per month. D. ​84.13% of cell phone plans cost less than ​$49 per month.

μ = -18 σ = 1

The graph of a normal curve is given on the right. Use the graph to identify the values of μ and σ.

μ = 22 σ = 3

The graph of a normal curve is given on the right. Use the graph to identify the values of μ and σ.

(a) The probability is 1/3 (b) There is a 50​% probability the friend will arrive within 15 minutes.

The graph to the right is the uniform probability density function for a friend who is x minutes late. ​(a) Find the probability that the friend is between 20 and 30 minutes late. ​(b) It is 10 A.M. There is a 50​% probability the friend will arrive within how many​ minutes?

(a) The proportion of human pregnancies that last more than 290 days is 0.0668. (b) The probability that a randomly selected human pregnancy lasts more than 290 days is 0.0668.

(a) The following figure represents the normal curve with μ = 266 days and σ = 16 days. The area to the right of X = 290 is 0.0668. Provide an interpretation of this area. A. The proportion of human pregnancies that last more than 290 days is 0.9332. B. The proportion of human pregnancies that last less than 290 days is 0.0668. C. The proportion of human pregnancies that last more than 290 days is 0.0668. Your answer is correct. D. The proportion of human pregnancies that last more than 266 days is 0.4332. (b) Provide a second interpretation of the area. A. The probability that a randomly selected human pregnancy lasts less than 290 days is 0.0668. B. The probability that a randomly selected human pregnancy lasts more than 290 days is 0.4332. C. The probability that a randomly selected human pregnancy lasts more than 290 days is 0.9332. D. The probability that a randomly selected human pregnancy lasts more than 290 days is 0.0668.

A graph could represent a normal density function if it is symmetric about its​ mean, it has a single peak at the​ mean, the highest point occurs at the​ mean, and if it​ approaches, but does not ​ reach, the horizontal axis as x increases without bound and decreases without bound. The function on the graph should take only positive values. This graph is not symmetric about its mean because it is skewed right. The graph has a single peak. The highest point occurs at the mean. This graph is always greater than or equal to zero. ​Thus, the graph cannot represent a normal density function because this graph is not symmetric.

Could the graph represent a normal density​ function?

This graph cannot represent a normal density function because this graph does not approach the x-axis as x increases or decreases.

Could the graph represent a normal density​ function?

No

Could the graph represent a normal density​ function? A. Yes B. No

This graph is symmetric about its mean. This graph approaches the​ x-axis as x increases or decreases without bound. This graph has three peaks.​ Thus, it is not​ bell-shaped. This graph cannot represent a normal density function because it has three peaks.

Determine whether the graph can represent a normal density function.

B

Draw a normal curve with μ = 55 and σ = 15. Label the mean and the inflection points.

Graph A has a mean of μ = 10 and graph B has a mean of μ = 14 because a larger mean shifts the graph to the right.

One graph in the figure represents a normal distribution with mean μ = 14 and standard deviation σ = 2. The other graph represents a normal distribution with mean μ = 10 and standard deviation σ = 2. Determine which graph is which and explain how you know. A. Graph A has a mean of μ = 10 and graph B has a mean of μ = 14 because a larger mean shifts the graph to the left. B. Graph A has a mean of μ = 14 and graph B has a mean of μ = 10 because a larger mean shifts the graph to the left. C. Graph A has a mean of μ = 14 and graph B has a mean of μ = 10 because a larger mean shifts the graph to the right. D. Graph A has a mean of μ = 10 and graph B has a mean of μ = 14 because a larger mean shifts the graph to the right.

*The statement is true. The normal curve is a symmetric distribution with one​ peak, which means the​ mean, median, and mode are all equal.​ Therefore, the normal curve is symmetric about the​ mean, μ.* For any symmetric​ distribution, the median equals the mean. The graph of the distribution must be symmetric about the median because it is the point where​ 50% of the area under the distribution lies on either side. The graph must also be symmetric about the​ mean, because it is the balancing point of a graph. For a symmetric​ graph, the balancing point will always be at the middle. The normal curve also only has one​ peak, and so for it to be​ symmetric, it must be symmetric at the highest point on the​ peak, which is the mode.​ Therefore, for the normal​ curve, mean = median = mode.

The normal curve is symmetric about its​ mean, μ. A.The statement is false. The normal curve is not symmetric about its​ mean, because the mean is the balancing point of the graph of the distribution. The median is the point where​ 50% of the area under the distribution is to the left and​ 50% to the right.​ Therefore, the normal curve could only be symmetric about its​ median, not about its mean. B. The statement is false. The mean is the balancing point for the graph of a​ distribution, and​ therefore, it is impossible for any distribution to be symmetric about the mean. C. The statement is true. The mean is the balancing point for the graph of a​ distribution, and​ therefore, all distributions are symmetric about the mean. D. The statement is true. The normal curve is a symmetric distribution with one​ peak, which means the​ mean, median, and mode are all equal.​ Therefore, the normal curve is symmetric about the​ mean, μ.

The points are x = μ - σ and x = μ + σ.

The points at x=​_______ and x=_______ are the inflection points on the normal curve. What are the two​ points? A. The points are x = μ - σ and x = μ + σ. B. The points are x = μ - 2σ and x = μ + 2σ. C.The points are x = μ - 3σ and x = μ + 3σ.