invNorm in
2nd DISTR . invNorm(area to the left, mean, standard deviation)
For this problem, invNorm(0.90,63,5) = 69.4
d. Find the 70 th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k ).
d. Find the 70 th percentile.
Draw a new graph and label it appropriately. k = 65.6
The 70 th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.
invNorm(0.70,63,5) = 65.6
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The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three.
Find the probability that a randomly selected golfer scored less than 65.
normalcdf(10 99 ,65,68,3) = 0.1587
A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.
a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.
a. Let X = the amount of time (in hours) a household personal computer is used for entertainment. X ~ N (2, 0.5) where μ = 2 and σ = 0.5.
Find P (1.8< x <2.75).
The probability for which you are looking is the area between x = 1.8 and x = 2.75. P (1.8< x <2.75) = 0.5886
normalcdf(1.8,2.75,2,0.5) = 0.5886
The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.
b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.
b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25 th percentile, k , where P ( x < k ) = 0.25.
invNorm(0.25,2,0.5) = 1.66
The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.
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The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.
normalcdf(66,70,68,3) = 0.4950
There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.
a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.
a. normalcdf(23,64.7,36.9,13.9) = 0.8186
b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.
b. normalcdf(–10
99 ,50.8,36.9,13.9) = 0.8413
c. Find the 80 th percentile of this distribution, and interpret it in a complete sentence.
c.
- invNorm(0.80,36.9,13.9) = 48.6
- The 80 th percentile is 48.6 years.
- 80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less.