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The following histogram shows the relative frequencies of the heights, recorded to the nearest inch, of a population of women. The mean of the population is 64.97 inches, and the standard deviation is 2.66 inches.
One woman from the population will be selected at random.

Question 3
(c) The histogram displays a discrete probability model for height. However, height is often considered a continuous variable that follows a normal model. Consider a normal model that uses the mean and standard deviation of the population of women as its parameters.

(i) Use the normal model and the relationship between area and relative frequency to find the probability that the randomly selected woman will have a height of at least 67 inches. Show your work.

(ii) Does your answer in part (c-i) match your answer in part (a) ? If not, give a reason for why the answers might be different.

Question 4
(d) Let the random variable H represent the height of a woman in the population. P(H<60) represents the probability of randomly selecting a woman with height less than 60 inches. Based on the information given, the probability can be found using either the discrete model or the normal model.

(i) Give an example of a probability of H that can be found using the discrete model but not the normal model. Explain why.

(ii) Give an example of a probability of H that can be found using the normal model but not the discrete model. Explain why.

The Following Histogram Shows The Relative Frequencies Of The Heights Recorded To The Nearest Inch Of A Population Of Women The Mean Of The Population Is 6497 I class=

Sagot :

Answer:

The following histogram shows the relative frequencies of the height recorded to the nearest inch of population of women the mean of the population is 64.97 inches and the standard deviation is 2.66 inches

(a) Based on the histogram, what is the probability that the selected woman will have a height of at least 67 inches? Show your work

Answer:

0.22268

Step-by-step explanation:

z-score is z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation.

(a) Based on the histogram, what is the probability that the selected woman will have a height of at least 67 inches? Show your work

At least means equal to or greater than 67 inches

z = 67 - 64.97/2.66

z = 0.76316

P-value from Z-Table:

P(x<67) = 0.77732

P(x>67) = 1 - P(x<67) = 0.22268

The probability that the selected woman will have a height of at least 67 inches is 0.22268

Step-by-step explanation:

A variable that can have any value between two given value is known as continuous variable, while a variable have only integer values is a discrete variable

The correct values are;

(c) (i) The probability of a height of at least 67 is 0.2263

(ii) The answers do not match due to the difference between the discrete and normal probability models

(d) (i) A probability that can be found using the discrete model but not the normal model is P (H = 60)

The probability of an exact value using the normal distribution is zero

(ii) A probability that can be found using the normal model but not the discrete model is P(H = 56)

The value is for height of 56 inches is not given on the histogram and therefore

The reason the above selections are correct is follows:

(c) (i) Using the normal probability model, we have;

The mean = 64.27 inches, the standard deviation = 2.66 inches

The z-score for the height of 67 inches is given as follows;

[tex]z = \mathbf{\dfrac{x - \mu}{\sigma}}[/tex]

Therefore, we have;

[tex]z = \dfrac{67 -64.97}{2.66} \approx 0.7632[/tex]

From the z-score table, the probability P(x < 67) = 0.7737

P(x > 67) = 1 - P(x < 67) = 1 - 0.7737 = 0.2263

The probability of a height of at least 67, by the normal model = 0.2263

(ii) Based on the histogram, we have;

The probability of at least 67 = The area of the bars equal to and larger than 67 = (0.11 + 0.07 + 0.05 + 0.02 + 0.01 + 0.01) × 1 = 0.27

The probability of a height of at least 67 = 0.27

The answer in part (c-i) and the answer in part (a) using the histogram do not match

The reason they do not match is that the normal model used in part c is

based on normal probability, which focuses on future events, using the

normal model which is a continuous probability model, while the value of

probability calculated using the histogram is based on the relative

frequency, which focuses on historical values, based on the discrete model

(d) (i) An example of a probability that can be found using the discrete model but not the normal model is the probability that the height of a woman is a discreet single value such as 65 inches

Therefore, using the discreet model, we have;

P(H = 65) = 0.18

Using the continuous model, we have;

P(H = 65) = 0

This is so because the continuous model is given by the area under the curve, which for a single value is infinitesimally small

(ii) An example of a probability that can be found using the normal model but not the discrete model is the probability of the height of a woman selected is 56 inches, P(H = 56) is not given in the histogram

Learn more about discrete and normal probability distributions here:

https://brainly.com/question/6476990

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