Question

A certain brand of light bulb is advertised to last 1600 hours. Suppose that the lifetimes of the light bulbs are normally distributed with a mean lifetime of 1650 hours and a standard deviation of 60 hours. (a) Sketch the normal distribution, go out 3 standard deviations.

(b) What proportion of the light bulbs last at least the advertised time of 1600 hours? P(x ≥ 1600) = (Use the normalcdf on your graphing calculator.)

(c) What proportion of light bulbs last less than 1500 hours? P(x < 1500) = (Use the normalcdf on your graphing calculator.)

(d) What proportion of light bulbs last between 1700 hours and 1800 hours? P(1700 < x < 1800) = (Use the normalcdf on your graphing calculator.)

(e) What is the 75th percentile of the distribution of light bulb lifetimes? (Use the invNorm function on your calculator.) “75% of the light bulbs last ___________ hours or less.

(f) What lifetimes make up the middle 90% of lifetimes? (Use the invNorm function on your calculator.) “90% of the light bulbs last between __________ hours and __________ hours.” (Hint: If the area of the middle region is 0.90, then the total area of the two tails is 1 – 0.90 = 0.10, and the area of each tail is ½(0.10) = 0.05. One of the boundaries is the 5th percentile of the distribution. The other boundary has an area of 0.05 to the right.)

Answer 1

Solution(a)

Solution(b)
Given Mean = 1650hours
Standard deviation = 60
P(X>=1600) = 1-P(X<1600)
Z = (1600-1650)/60 = -50/60 = -0.8333
from Z table we can find out p-value
P(X>=1600) = 1- 0.2033 = 0.7967 or 79.67%
Solution(c)
P(X<1500)=?
Z = (1500-1650)/60 = -150/60 = -2.5
P(X<1500) = 0.0062
Solution(d)
P(1700<x<1800)= P(X<1800)-P(X<1700)
Z = 1800-1650/60 = 2.5
Z = 1700-1650/60 = 0.833
P(1700<x<1800) = 0.9938 - 0.7967 = 0.1971
Solution(e)
p=0.75 and Z score = 0.675
0.675 = (X-1650)/60
X = 1650+40.5 = 1690.5
Solution(f)
alpha = 0.10 and p-value = 0.05 it is two tailed test
Z = +/-1.645
1.645 = (X-1650)/60
Xbar = 1650+98.7 = 1748.7
-1.645 = (X-1650)/60
X = 1650-98.7 = 1551.3