Respuesta :
Answer:
a)3.438% of the light bulbs will last more than 6262 hours.
b)11.31% of the light bulbs will last 5252 hours or less.
c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.
d) 0.12% of the light bulbs will last 4646 hours or less.
Step-by-step explanation:
Normally distributed problems can be solved by the z-score formula:
On a normaly distributed set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a value X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.
In this problem, we have that:
The lifetimes of light bulbs are approximately normally distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.
So [tex]\mu = 5656, \sigma = 333.3[/tex]
(a) What proportion of light bulbs will last more than 6262 hours?
The pvalue of the z-score of [tex]X = 6262[/tex] is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{6262 - 5656}{333.3}[/tex]
[tex]Z = 1.82[/tex]
[tex]Z = 1.81[/tex] has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So
[tex]P = 100% - 96.562% = 3.438%[/tex] of the light bulbs will last more than 6262 hours.
(b) What proportion of light bulbs will last 5252 hours or less?
This is the pvalue of the zscore of [tex]X = 5252[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5252- 5656}{333.3}[/tex]
[tex]Z = -1.21[/tex]
[tex]Z = -1.21[/tex] has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.
(c) What proportion of light bulbs will last between 5858 and 6262 hours?
This is the pvalue of the zscore of [tex]X = 6262[/tex] subtracted by the pvalue of the zscore [tex]X = 5858[/tex]
For [tex]X = 6262[/tex], we have that [tex]Z = 1.81[/tex] with a pvalue of .96562.
For X = 5858
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5858- 5656}{333.3}[/tex]
[tex]Z = 0.61[/tex]
[tex]Z = 0.61[/tex] has a pvalue of .72907.
So, the proportion of light bulbs that will last between 5858 and 6262 hours is
[tex]P = .96562 - .72907 = .23655[/tex]
23.655% of the light bulbs are going to last between 5858 and 6262 hours.
(d) What is the probability that a randomly selected light bulb lasts less than 4646 hours?
This is the pvalue of the zscore of [tex]X = 4646[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4646- 5656}{333.3}[/tex]
[tex]Z = -3.03[/tex]
[tex]Z = -3.03[/tex] has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.
