Respuesta :
Answer:
0.9864
Step-by-step explanation:
The average sample mean is 295, and the standard deviation is:
σ = 15 / √17
σ = 3.64
The z-scores are:
z = (x − μ) / σ
z₁ = (-9) / 3.64
z₁ = -2.47
z₂ = (9) / 3.64
z₂ = 2.47
So the probability is:
P(-2.47 < Z < 2.47) = P(Z < 2.47) − P(Z < -2.47)
P(-2.47 < Z < 2.47) = 0.9932 − 0.0068
P(-2.47 < Z < 2.47) = 0.9864
Using the normal distribution and the central limit theorem, it is found that there is a 0.9864 = 98.64% probability that a random sample of 17 pregnancies has a mean gestation period within 9 days of the mean.
In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- For a sample of size n, by the Central Limit Theorem, the standard deviation is given by [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem:
- Mean of 295 days, thus [tex]\mu = 295[/tex].
- Standard deviation of 15 days, thus [tex]\sigma = 15[/tex].
- Sample of 17 pregnancies, thus [tex]n = 17, s = \frac{15}{\sqrt{17}}[/tex].
Gestation period within 9 days of the mean is between 286 and 304 days, and the probability is the p-value of Z when X = 304 subtracted by the p-value of Z when X = 286.
X = 304:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{304 - 295}{\frac{15}{\sqrt{17}}}[/tex]
[tex]Z = 2.47[/tex]
[tex]Z = 2.47[/tex] has a p-value of 0.9932
X = 286:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{286 - 295}{\frac{15}{\sqrt{17}}}[/tex]
[tex]Z = -2.47[/tex]
[tex]Z = -2.47[/tex] has a p-value of 0.0068
0.9932 - 0.0068 = 0.9864.
0.9864 = 98.64% probability that a random sample of 17 pregnancies has a mean gestation period within 9 days of the mean.
A similar problem is given at https://brainly.com/question/24663213
