standard normal distribution percentages

Forty percent of the ages that range from 13 to 55+ are at least what age?.

Calculate the interquartile range ( IQR).

Using this information, answer the following questions (round answers to one decimal place). 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. There are approximately one billion smartphone users in the world today. TI-Calculators: invNorm(0.30, 36.9, 13.9)ģ0% of smartphone users aged between 13 and 55+ are at most 29.6 years and 70% are at least 29.6 years. The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.

To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, we need to find the 25th percentile, k.

The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886. X ~ N(2, 0.5) where mean, μ = 2 and standard deviation, σ = 0.5.įind P(1.8 < x < 2.75).The probability for which you are looking is the area between x = 1.8 and x = 2.75.

Let X= the amount of time (in hours) a household personal computer is used for entertainment.

Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.

Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.

Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. The proability that a randomly selected golfer scored less than 65 is 0.1587.Ī personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Find the 70th percentile.Draw a new graph and label it appropriately.This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. The format for this function is invNorm(area to the left, mean, standard deviation)įor this problem, we need to input 'invNorm(0.90,63,5)' and press '='. Shade the area that corresponds to the 90th percentile. Find the 90th percentile.For each problem or part of a problem, draw a new graph.ĭraw the x-axis.The probability that one student scores less than 85 is approximately one (or 100%). The number –10 99 is way out in the left tail of the normal curve. In some instances, the lower number of the area might be –10 99. We are calculating the area when final exam score is between between 65 and 10 99. The number 10 99 is way out in the right tail of the normal curve. The probability that any student selected at random scores more than 65 is 0.3446. The syntax for the instructions are as follows: normalcdf (lower value, upper value, mean, standard deviation).įor this problem: normalcdf (65,10^99,63,5) and press '='. X ~ N(63, 5), where mean, μ = 63 and standard deviation,σ = 5. Find the probability that a randomly selected student scored more than 65 on the exam.Let X = a score on the final exam.

Find the 70th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k).Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k).

Find the probability that a randomly selected student scored less than 85.Find the probability that a randomly selected student scored more than 65 on the exam.The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of 5. Example 2 (by using TI-83/84 Calculators)

Standard normal distribution percentages