Example

The final exam scores in a statistics course were normally spread with a mean of 63 and a traditional deviation the five.

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Find the probability that a randomly selected college student scored more than 65 ~ above the exam.Find the probability that a randomly selected college student scored much less than 85.Find the 90th percentile (that is, find the score k that has actually 90% that the scores below k and also 10% of the scores over k).Find the 70th percentile (that is, discover the score k such that 70% of scores are below k and 30% of the scores are above k).

Solution:

LetX = a score on the last exam. X ~ N(63, 5), whereby μ = 63 and also σ = 5Draw a graph. Then, uncover P(x > 65). P(x > 65) = 0.3446
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The probability that any student selected at random scores more than 65 is 0.3446.

Go into second DISTR. After pressing 2nd DISTR, press2:normalcdf. The syntax because that the instructions are as follows:normalcdf(lower value, top value, mean, standard deviation) because that this problem: normalcdf(65,1E99,63,5) = 0.3446. You obtain 1E99 (= 1099) by pushing 1, the EE crucial (a 2nd key) and then 99. Or, you deserve to enter10^99 instead. The number 1099 is means out in the ideal tail the the common curve. We are calculating the area between 65 and 1099. In some instances, the lower variety of the area can be –1E99 (= –1099). The number –1099 is way out in the left tail of the typical curve.displaystylez=frac65-635=0.4Area come the left is 0.6554.P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446Calculate the z-score:*Press2nd Distr*Press 3:invNorm(*Enter the area to the left the z complied with by )*Press ENTER.For this Example, the procedures are2nd Distr3:invNorm(.6554) ENTERThe price is 0.3999 which ring to 0.4.Draw a graph. Then find P(x find the 90th percentile. For each problem or part of a problem, attract a new graph. Draw the x-axis. The shade the area that coincides to the 90th percentile.Let k = the 90th percentile. The change k is situated on the x-axis. P(x
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k = 65.6The 70th percentile is 65.6. This means that 70% of the check scores loss at or listed below 65.5 and 30% fall at or above.invNorm(0.70,63,5) = 65.6

Example

A personal computer is provided for office work-related at home, research, communication, an individual finances, education, entertainment, social networking, and also a myriad of various other things. Suppose that the average variety of hours a household an individual computer is offered for entertain is two hours per day. Assume the times for entertainment are generally distributed and the conventional deviation because that the times is half an hour.

Find the probability the a household personal computer is offered for entertainment between 1.8 and 2.75 hours per day.Find the maximum variety of hours per day that the bottom quartile of family members uses a an individual computer because that entertainment.

Solution:

Let X= the quantity of time (in hours) a household an individual computer is supplied for entertainment. X ~ N(2, 0.5) wherein μ = 2 and also σ = 0.5. Discover P(1.8 betweenx = 1.8 and x = 2.75. P(1.8
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normalcdf(1.8,2.75,2,0.5) = 0.5886The probability that a household personal computer is used between 1.8 and also 2.75 hrs per day because that entertainment is 0.5886.To uncover the maximum variety of hours per day that the bottom quartile of family members uses a personal computer because that entertainment, find the 25th percentile, k, whereby P(x
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invNorm(0.25,2,0.5) = 1.66The maximum number of hours per day the the bottom quartile of families uses a an individual computer because that entertainment is 1.66 hours.

try it

Use the info in instance 3 come answer the adhering to questions.

Find the 30th percentile, and interpret the in a complete sentence.What is the probability the the age of a randomly selected smartphone user in the selection 13 come 55+ is less than 27 year old.

LetX = a clever phone user whose age is 13 come 55+. X ~ N(36.9, 13.9)

To uncover the 30th percentile, findk such the P(x find P(x
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Example

A citrus farmer that grows mandarin oranges finds the the diameters of mandarin oranges harvested on his farm follow a normal distribution with a average diameter the 5.85 cm and also a traditional deviation of 0.24 cm.

Find the probability that a randomly selected mandarin orange from this farm has actually a diameter larger than 6.0 cm. Map out the graph.The middle 20% the mandarin oranges from this farm have actually diameters in between ______ and also ______.Find the 90th percentile for the diameters the mandarin oranges, and also interpret that in a complete sentence.

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Solution:

normalcdf(6,10^99,5.85,0.24) = 0.2660
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1 – 0.20 = 0.80 The tails of the graph that the normal distribution each have actually an area the 0.40. Uncover k1, the 40th percentile, and also k2, the 60th percentile (0.40 + 0.20 = 0.60). k1 = invNorm(0.40,5.85,0.24) = 5.79 centimeter k2 = invNorm(0.60,5.85,0.24) = 5.91 cm6.16: Ninety percent that the diameter the the mandarin oranges is at most 6.15 cm.