The following histogram shows the relative frequencies of the heights, recorded to the nearest inch, of a population of women. T
he 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.
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.
1 answer:
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Answer:
0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
In this problem, we have that:
What is the probability that the sample mean would differ from the true mean by less than 1.11 months?
This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So
X = 120.1
By the Central Limit Theorem
has a pvalue of 0.7517
X = 117.9
has a pvalue of 0.2483
0.7517 - 0.2483 = 0.5034
0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months