Respuesta :
Using the normal distribution, we have that:
a) 0.2266 = 22.66% of pregnancies lasts more than 280 days.
b) 0.4649 = 46.49% of pregnancies lasts between 256 and 276 days.
c) 0.0401 = 4.01% probability that a randomly selected pregnancy lasts no more than 240 days.
d) |Z| < 2, hence very preterm babies are not unusual.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
- If |Z| > 2, the measure is considered unusual.
The mean and the standard deviation are given as follows:
[tex]\mu = 268, \sigma = 16[/tex].
Item a:
The proportion is one subtracted by the p-value of Z when X = 280, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{280 - 268}{16}[/tex]
Z = 0.75
Z = 0.75 has a p-value of 0.7734.
1 - 0.7734 = 0.2266.
0.2266 = 22.66% of pregnancies lasts more than 280 days.
Item b:
The proportion is the p-value of Z when X = 276 subtracted by the p-value of Z when X = 256, hence:
X = 276:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{276 - 268}{16}[/tex]
Z = 0.5
Z = 0.5 has a p-value of 0.6915.
X = 256:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{256 - 268}{16}[/tex]
Z = -0.75
Z = -0.75 has a p-value of 0.2266.
0.6915 - 0.2266 = 0.4649.
0.4649 = 46.49% of pregnancies lasts between 256 and 276 days.
Item c:
The probability is the p-value of Z when X = 240, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{240 - 268}{16}[/tex]
Z = -1.75
Z = -1.75 has a p-value of 0.0401.
0.0401 = 4.01% probability that a randomly selected pregnancy lasts no more than 240 days.
Item d:
We have to find the z-score.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{244 - 268}{16}[/tex]
Z = -1.5.
|Z| < 2, hence very preterm babies are not unusual.
More can be learned about the normal distribution at https://brainly.com/question/4079902
#SPJ1
