They had conscripted men to form a colonial army is about the colonies in April 1775.
What happened in the colonies in 1775?Tensions between citizens of the 13 North American colonies of Great Britain and the colonial administration, which represented the British crown, grew over time, leading to the Revolutionary War (1775–83), also known as the American Revolution.Successive generations have come to refer to those events as the "shot heard round the world." The battle on April 19, 1775, which is frequently referred to as the "Battles of Lexington and Concord," lasted across 16 miles along the Bay Road from Boston to Concord and involved roughly 4,000 Americans and 1,700 British regulars.Boston, Massachusetts | 19 April 1775 Massachusetts colonists disobeyed British rule in the opening fight of the American Revolution, outnumbered and outfought the Redcoats, and began a protracted war to obtain their freedom.To learn more about American colonies refer to:
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Assume that the duration of human pregnancies can be described by a Normal model with mean
265 days and standard deviation 13 days.
a) What percentage of pregnancies should last between 269 and
277 days?
b) At least how many days should the longest 25% of all pregnancies last?
c) Suppose a certain obstetrician is currently providing prenatal care to 55 pregnant patients. Let
y represent the mean length of their pregnancies. According to the Central Limit Theorem, what is the distribution of this sample mean, y? Specify the model, mean, and standard deviation.
d) What's the probability that the mean duration of these patients' pregnancies will be less than
254 days?
a) The percentage of pregnancies that should last between 269 and 277 days is of: 19.95%.
b) The longest 25% of pregnancies should last at least 274 days.
c) The distribution of y is approximately normal with mean of 265 days and standard deviation of 1.75 days.
d) The probability that the mean duration of these patients' pregnancies will be less than 254 days is of 0.
How to obtain probabilities using the normal distribution?The z-score of a measure X of a variable that has mean symbolized by [tex]\mu[/tex] and standard deviation symbolized by [tex]\sigma[/tex] is obtained by the rule presented as follows:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The z-score represents how many standard deviations the measure X is above or below the mean of the distribution, depending if the obtained z-score is positive or negative.Using the z-score table, the p-value associated with the calculated z-score is found, and it represents the percentile of the measure X in the distribution.By the Central Limit Theorem, the sampling distribution of sample means of size n is approximately normal with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].The mean and the standard deviation of the pregnancies lengths in this problem are given as follows:
[tex]\mu = 265, \sigma = 13[/tex]
The proportion of pregnancies that last between 269 and 277 days is the p-value of Z when X = 277 subtracted by the p-value of Z when X = 269, hence:
X = 277:
Z = (277 - 265)/13
Z = 0.92
Z = 0.92 has a p-value of 0.8212.
X = 269:
Z = (269 - 265)/13
Z = 0.31
Z = 0.31 has a p-value of 0.6217.
Hence:
0.8212 - 0.6217 = 0.1995 = 19.95%.
The longest 25% of pregnancies last at least the 75th percentile, which is X when Z = 0.675, hence:
0.675 = (X - 265)/13
X - 265 = 0.675 x 13
X = 274 days.
For samples of 55, the standard deviation is given as follows:
s = 13/sqrt(55) = 1.75 days.
The probability that the mean duration of these patients' pregnancies will be less than 254 days is the p-value of Z when X = 254, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
Z = (254 - 265)/1.75
Z = -6.29
Z = -6.29 has a p-value of 0.
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