Respuesta :
Answer:
1. 43.32% of bulbs will have a life between 1350 and 1377 hr.
2. 19.15% of bulbs will have a life between 1341 and 1350 hr.
3. 54.53% of bulbs will have a life between 1338 and 1365 hr.
4. 13.65% of bulbs will have a life between 1365 and 1377 hr.
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 = 1350, \sigma = 18[/tex]
1. What percentage of bulbs will have a life between 1350 and 1377 hr?
This is the pvalue of Z when X = 1377 subtracted by the pvalue of Z when X = 1350.
X = 1377
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1377 - 1350}{18}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a pvalue of 0.9332.
X = 1350
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1350 - 1350}{18}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a pvalue of 0.5.
So 0.9332 - 0.5 = 0.4332 = 43.32% of bulbs will have a life between 1350 and 1377 hr.
2. What percentage of bulbs will have a life between 1341 and 1350 hr?
This is the pvalue of Z when X = 1350 subtracted by the pvalue of Z when X = 1341.
X = 1350
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1350 - 1350}{18}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a pvalue of 0.5.
X = 1341
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1341 - 1350}{18}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] has a pvalue of 0.3085.
So 0.5 - 0.3085 = 0.1915 = 19.15% of bulbs will have a life between 1341 and 1350 hr.
3. What percentage of bulbs will have a life between 1338 and 1365 hr?
This is the pvalue of Z when X = 1365 subtracted by the pvalue of Z when X = 1338.
X = 1365
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1365 - 1350}{18}[/tex]
[tex]Z = 0.83[/tex]
[tex]Z = 0.83[/tex] has a pvalue of 0.7967
X = 1338
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1338 - 1350}{18}[/tex]
[tex]Z = -0.67[/tex]
[tex]Z = -0.67[/tex] has a pvalue of 0.2514.
So 0.7967 - 0.2514 = 0.5453 = 54.53% of bulbs will have a life between 1338 and 1365 hr.
4. What percentage of bulbs will have a life between 1365 and 1377 hr?
This is the pvalue of Z when X = 1377 subtracted by the pvalue of Z when X = 1365.
X = 1377
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1377 - 1350}{18}[/tex]
[tex]Z = 1.5[/tex]
[tex]Z = 1.5[/tex] has a pvalue of 0.9332.
X = 1365
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1365 - 1350}{18}[/tex]
[tex]Z = 0.83[/tex]
[tex]Z = 0.83[/tex] has a pvalue of 0.7967.
So 0.9332 - 0.7967 = 0.1365 = 13.65% of bulbs will have a life between 1365 and 1377 hr.
