Respuesta :
Answer:
c. 0.136.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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.
In this problem, we have that:
[tex]\mu = 110, \sigma = 15[/tex]
The proportion of infants with birth weights between 125 oz and 140 oz is
This is the pvalue of Z when X = 140 subtracted by the pvalue of Z when X = 125. So
X = 140
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{140 - 110}{15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.977
X = 125
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{125 - 110}{15}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.841
0.9772 - 0.841 = 0.136
So the correct answer is:
c. 0.136.
Answer: c. 0.136.
Step-by-step explanation:
Since the Birth weights at a local hospital have a Normal distribution,
we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = Birth weights.
µ = mean weight
σ = standard deviation
From the information given,
µ = 110 oz
σ = 15 oz
We want to find the proportion of infants with birth weights between 125 oz and 140 oz. It is expressed as
P(125 ≤ x ≤ 140)
For x = 125,
z = (125 - 110)/15 = 1
Looking at the normal distribution table, the probability corresponding to the z score is 0.841
For x = 140,
z = (140 - 110)/15 = 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.977
Therefore,
P(125 ≤ x ≤ 140) = 0.977 - 0.841 = 0.136
