<< Chapter < Page Chapter >> Page >

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

Got questions? Get instant answers now!

Try it

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

Got questions? Get instant answers now!

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

This is a normal distribution curve. The peak of the curve coincides with the point 2 on the horizontal axis. The values 1.8 and 2.75 are also labeled on the x-axis. Vertical lines extend from 1.8 and 2.75 to the curve. The area between the lines is shaded.

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.

Got questions? Get instant answers now!

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.

This is a normal distribution curve. The area under the left tail of the curve is shaded. The shaded area shows that the probability that x is less than k is 0.25. It follows that k = 1.67.

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.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Try it

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

Got questions? Get instant answers now!

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

Got questions? Get instant answers now!

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

Got questions? Get instant answers now!

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.

Got questions? Get instant answers now!
Got questions? Get instant answers now!

Questions & Answers

binomial distribution tends to normal distribution
Murali Reply
if sample size n very large and probability tends to 0.5 if these both conditions are satisfied by binomial distribution it would tends to normal distribution
sravani
n tends to infinite i.e large Probability tends to 0 i.e indefinitely small. np = lamda
Anji
the above is poison to Bin
Anji
no of trails n tends to indefinitely large..i.e infine neither p nor q is very small Then bin tends to normal
Anji
if the death of of the snow is my yard is normally distributed with the m is equals to 2.5 and what is the probability that a randomly chosen location with have a no that between 2.25 and 2.76
Sakshi Reply
hey
Shubham
🤔
Iqra
hello
Sakshi
hii
Rushikesh
helow
why Statistics so hard
Mohd
ho geya solve
Sakshi
it's not hard
Sakshi
it is hard 😭
Mohd
solution?
Abdul
hii
Aadil
it's just need to be concentrate
Akinyemi
exactly..... concentration is very important
Iqra
rewrite the question
Aadil
what is the true statement about random variable?
Henna Reply
A consumer advocate agency wants to estimate the mean repair cost of a washing machine. the agency randomly selects 40 repair cost and find the mean to be $100.00.The standards deviation is $17.50. Construct a 90% confidence interval for the mean.
Deshah Reply
pls I need understand this statistics very will is giving me problem
Bolanle Reply
Sixty-four third year high school students were given a standardized reading comprehension test. The mean and standard deviation obtained were 52.27 and 8.24, respectively. Is the mean significantly different from the population mean of 50? Use the 5% level of significance.
Daryl Reply
No
Ariel
how do I find the modal class
Bruce Reply
look for the highest occuring number in the class
Kusi
the probability of an event occuring is defined as?
James Reply
The probability of an even occurring is expected event÷ event being cancelled or event occurring / event not occurring
Gokuna
what is simple bar chat
Toyin Reply
Simple Bar Chart is a Diagram which shows the data values in form of horizontal bars. It shows categories along y-axis and values along x-axis. The x-axis displays above the bars and y-axis displays on left of the bars with the bars extending to the right side according to their values.
Muhammad
statistics is percentage only
Moha Reply
the first word is chance for that we use percentages
muhammad
it is not at all that statistics is a percentage only
Shambhavi
I need more examples
Luwam Reply
how to calculate sample needed
Jim Reply
mole of sample/mole ratio or Va Vb
Gokuna
how to I solve for arithmetic mean
Joe Reply
Yeah. for you to say.
James
yes
niharu
how do I solve for arithmetic mean
Joe Reply
please answer these questions
niharu
add all the data and divide by the number of data sets. For example, if test scores were 70, 60, 70, 80 the total is 280 and the total data sets referred to as N is 4. Therfore the mean or arthritmatic average is 70. I hope this helps.
Jim
*Tan A - Tan B = sin(A-B)/CosA CosB ... *2sinQ/Cos 3Q = tan 3Q - tan Q
Ibraheem Reply

Get the best Introductory statistics course in your pocket!





Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Introductory statistics' conversation and receive update notifications?

Ask