Assume that the duration of human pregnancies can be described by a Normal model with mean 266 days and standard deviation 16 days. a) What percentage of pregnancies should last between 270 and 280 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 60 pregnant women. Let y̅ represent the mean length of their pregnancies. According to the Central Limit Theorem, what's the distribution of this sample mean, y̅? Specify the model, mean, and standard deviation. d) What's the probability that the mean duration of these patient's pregnancies will be less than 260 days?

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Assume that the duration of human pregnancies can be described by a Normal model with mean 266 days and standard deviation 16 days. a) What percentage of pregnancies should last between 270 and 280 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 60 pregnant women. Let y̅ represent the mean length of their pregnancies. According to the Central Limit Theorem, what&#039;s the distribution of this sample mean, y̅? Specify the model, mean, and standard deviation. d) What&#039;s the probability that the mean duration of these patient&#039;s pregnancies will be less than 260 days?

enhebrevz · Accepted Answer

a) Soo we have &amp;nbsp;μ=266&amp;nbsp; and &amp;nbsp;σ=16&amp;nbsp;. Hence,P(270≤X≤280)=P(270−26616≤z≤280−26616)=P(0.25≤z≤0.88)&amp;nbsp;P(0.25≤z≤0.88)=P(z≤0.88)−P(z≤0.25)&amp;nbsp;=0.8106−0.5987&amp;nbsp;=0.2119&amp;nbsp;So, percentage will be 21.1%&amp;nbsp;b)&amp;nbsp;P(Z≥z)=0.25&amp;nbsp;We know z=0.675&amp;nbsp;x−26616=0.675&amp;nbsp;x=276.8&amp;nbsp;So andwer is 277 days.&amp;nbsp;c) We know&amp;nbsp;μ=266&amp;nbsp;σ=1660=2.06&amp;nbsp;d)&amp;nbsp;P(X&amp;lt;260)=P(z&amp;lt;260−2662.06)=P(z&amp;lt;−2.914)=0.00187&amp;nbsp;So, answer is is 0.00187

Don Sumner · Accepted Answer

a) The percentage of pregnancies that should last between 270 and 280 days should be determined, we can use the properties of the Normal distribution. Let X represent the duration of pregnancies, which follows a Normal distribution with a mean of μ=266 days and a standard deviation of σ=16 days.We want to find P(270≤X≤280). Converting this to standard units using the z-score formula, we have:P(270−μσ≤X−μσ≤280−μσ)Substituting the values, we obtain:P(270−26616≤X−26616≤280−26616)Simplifying, we have:P(0.25≤X−26616≤0.875)Using standard Normal tables or a calculator, we can find the corresponding probabilities:P(0.25≤Z≤0.875)≈P(Z≤0.875)−P(Z≤0.25)Consulting the standard Normal table or using a calculator, we find P(Z≤0.875)≈0.8078 and P(Z≤0.25)≈0.5987.Therefore, the percentage of pregnancies that should last between 270 and 280 days is approximately:0.8078−0.5987=0.2091&amp;nbsp;or&amp;nbsp;20.91%b) To find the number of days the longest 25% of all pregnancies should last, we can use the concept of the inverse Normal distribution. We need to find the value x such that P(X&amp;gt;x)=0.25, where X follows a Normal distribution with μ=266 days and σ=16 days.Using the inverse Normal table or a calculator, we find P(Z&amp;gt;−0.674)=0.25. Therefore, we have:X−26616&amp;gt;−0.674Solving for X, we find:X&amp;gt;−0.674×16+266Calculating, we obtain:X&amp;gt;254.184Rounding up to the nearest whole number, we conclude that the longest 25% of all pregnancies should last at least 255 days.c) According to the Central Limit Theorem (CLT), when a sample is drawn from a population with any distribution, as the sample size increases, the distribution of the sample mean approaches a Normal distribution. In this case, the sample mean of the length of pregnancies, y¯, will follow a Normal distribution.The model of the sample mean, y¯, is a Normal distribution with a mean equal to the population mean, which is μ=266 days. The standard deviation of the sample mean, y¯, can be calculated using the formula σy¯=σn, where σ is the population standard deviation and n is the sample size. In this case, σ represents the population standard deviation, which is given as 16 days. And n represents the sample size, which is stated as 60 pregnant women in the question.Thus, the distribution of the sample mean, denoted by y¯, is a Normal distribution with a mean equal to the population mean (μ) of 266 days and a standard deviation given by:σy¯=σn=1660.Therefore, the model of the sample mean y¯ is a Normal distribution with a mean of 266 days and a standard deviation of 1660 days.d) To find the probability that the mean duration of these patients&#039; pregnancies will be less than 260 days, we can use the properties of the Normal distribution and the sample mean distribution derived in part c.We want to find P(y¯&amp;lt;260), where y¯ follows a Normal distribution with a mean of 266 days and a standard deviation of 1660 days.To calculate this probability, we need to standardize the variable y¯ using the z-score formula:z=y¯−μy¯σy¯=260−2661660.Simplifying, we have:z=−61660=−66016.Using the standard Normal table or a calculator, we can find the probability P(z&amp;lt;−66016).Evaluating this probability, we find:P(z&amp;lt;−66016)≈0.0002.Therefore, the probability that the mean duration of these patients&#039; pregnancies will be less than 260 days is approximately 0.0002, or 0.02%.

user_27qwe · Accepted Answer

a) To find the percentage of pregnancies that should last between 270 and 280 days, we can calculate the z-scores for both values and use the standard normal distribution.The z-score for 270 days is given by:z1=270−26616The z-score for 280 days is given by:z2=280−26616Using a standard normal distribution table or calculator, we can find the corresponding probabilities:P(270≤X≤280)=P(z1≤270−26616)−P(z2≤280−26616)b) To determine the number of days the longest 25% of pregnancies should last, we need to find the z-score that corresponds to the upper 25th percentile of the standard normal distribution. This can be calculated as:z=invNorm(0.75)To obtain the actual number of days, we can use the z-score formula:x=mean+(z×standard&amp;nbsp;deviation)c) According to the Central Limit Theorem, when the sample size is sufficiently large, the distribution of the sample mean follows a normal distribution. In this case, the distribution of the sample mean, y¯, will also be approximately normal.The model for the distribution of the sample mean y¯ is:y¯~Normal(μ,σn)where μ is the mean of the population (266 days), σ is the standard deviation of the population (16 days), and n is the sample size (60 pregnant women).d) To find the probability that the mean duration of the 60 patients&#039; pregnancies will be less than 260 days, we can calculate the z-score for 260 days using the formula:z=260−meanstandard&amp;nbsp;deviationnThen, we can use the standard normal distribution to find the corresponding probability:P(y¯&amp;lt;260)=P(z&amp;lt;260−meanstandard&amp;nbsp;deviationn)

RizerMix · Accepted Answer

a) By calculating the z-scores for the two values and then using the typical normal distribution table, it is possible to determine the proportion of pregnancies that should last between 270 and 280 days. The formula for the z-score is z=x−μσ, where x is the given value, μ is the mean, and σ is the standard deviation.For 270 days:z1=270−26616=0.25For 280 days:z2=280−26616=0.875Using the standard normal distribution table, we can find the corresponding probabilities:P(270&amp;lt;x&amp;lt;280)=P(0.25&amp;lt;z&amp;lt;0.875)Looking up these values in the table, we find:P(0.25&amp;lt;z&amp;lt;0.875)=0.3162−0.5987=0.2825Therefore, the percentage of pregnancies that should last between 270 and 280 days is 28.25%.b) To find the number of days at least required for the longest 25% of all pregnancies, we need to calculate the z-score corresponding to the 75th percentile of the standard normal distribution.Using the standard normal distribution table, we find the z-score corresponding to the 75th percentile is approximately 0.674.Now, we can use the z-score formula to find the corresponding value in terms of days:0.674=x−μσSolving for x, we get:x=0.674×16+266x≈277.984Therefore, the longest 25% of all pregnancies should last at least 278 days.c) According to the Central Limit Theorem, the distribution of the sample mean y¯ is approximately a normal distribution. The mean of the sample mean is equal to the population mean, and the standard deviation of the sample mean is given by the population standard deviation divided by the square root of the sample size. In this case, the sample size is 60.Therefore, the distribution of the sample mean y¯ can be represented as:y¯~N(μ,σn)where μ is the population mean (266 days) and σ is the population standard deviation (16 days), and n is the sample size (60).d) To find the probability that the mean duration of these patients&#039; pregnancies will be less than 260 days, we can calculate the z-score for the value 260 using the sample mean distribution formula:z=x¯−μσnSubstituting the given values:z=260−2661660Simplifying:z=−62.066z≈−2.903Using the standard normal distribution table, we can find the probability associated with this z-score:P(x¯&amp;lt;260)=P(z&amp;lt;−2.903)From the table, we find:P(z&amp;lt;−2.903)≈0.0019Therefore, the probability that the mean duration of these patients&#039; pregnancies will be less than 260 days is approximately 0.19%.

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