Respuesta :
Answer:
B. 0.0918
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 = 18, \sigma = 1.5[/tex]
What proportion of batteries would be expected to last less than 16 hours?
This is the pvalue of Z when X = 16. So:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16 - 18}{1.5}[/tex]
[tex]Z = -1.33[/tex]
[tex]Z = -1.33[/tex] has a pvalue of 0.0918.
So the correct answer is:
B. 0.0918
Answer: the correct option is B
Step-by-step explanation:
the durations of the batteries are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = durations of the batteries in hours
u = mean time
s = standard deviation
From the information given,
u = 18 hours
s = 1.5 hours
We want to find the proportion or probability of batteries would be expected to last less than 16 hours. It is expressed as
P(x lesser than 16)
For x = 16,
z = (16 - 18)/1.5 = - 1.33
Looking at the normal distribution table, the probability corresponding to the z score is 0.09176
Approximately 0.0918
