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
betweenx = 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.
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.
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.
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
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.
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.
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.
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