Intro to stats 8,9,10

An English professor assigns letter grades on a test according to the following scheme. A: Top 13% of scores B: Scores below the top 13% and above the bottom 61% C: Scores below the top 39% and above the bottom 20% D: Scores below the top 80% and above the bottom 10% F: Bottom 10% of scores Scores on the test are normally distributed with a mean of 72.8 and a standard deviation of 7.3. Find the numerical limits for a D grade. Round your answers to the nearest whole number, if necessary.

invNorm (.20, 72.8, 7.3) invNorm (.10, 72.8, 7.3)

Find the value of z such that 0.9108 of the area lies between −z and z

invNorm ((1-.9108)/2, 0, 1)

Find the value of z such that 0.07 of the area lies to the left of z.

invnorm(.07,0,1) make negative if it says to the left

Suppose the horses in a large stable have a mean weight of 873lbs, and a variance of 17,161. What is the probability that the mean weight of the sample of horses would differ from the population mean by greater than 15lbs if 50 horses are sampled at random from the stable?

normal cdf( 873+15, 10000, 873, sqrtvar/sqrt50)X2

19% defectives. If a sample of size 303 is selected, what is the probability that the sample proportion will be greater than 17%? Round your answer to four decimal places.

normalcdf ( .17, 1000, .19, sqrt p(1-p)/367)

The mean per capita income is 19,695 dollars per annum with a variance of 802,816. What is the probability that the sample mean would differ from the true mean by less than 158 dollars if a sample of 226 persons is randomly selected? Round your answer to four decimal places.

normalcdf ( 19695-158, 19695+158, 19695, sqrtvar/sqrtsample)

Suppose the true proportion is 0.05. If 304 are sampled, what is the probability that the sample proportion will differ from the population proportion by more than 0.03? Round your answer to four decimal places.

normalcdf (.05+.03, 100, .05, sqrt p(1-p)/304) X 2

Suppose the true proportion is 0.06. If 367 are sampled, what is the probability that the sample proportion will be less than 0.08? Round your answer to four decimal places.

normalcdf (0, .08, .06, sqrt p(1-p)/367)

Trucks in a delivery fleet travel a mean of 120 miles per day with a standard deviation of 19 miles per day. The mileage per day is distributed normally. Find the probability that a truck drives less than 146 miles in a day

normalcdf (0, 146, 120, 19)

Find the area under the standard normal curve between z=2.66 and z=2.8

normalcdf (2.66, 2.8, 0, 1)

Calculate the z-score of the given X value, X=52.6, where μ=48.9 and σ=48.4 and indicate on the curve where z will be located

(x-u)/standard deviation (52.6-48.9)/48.4

A particular employee arrives at work sometime between 8:00 a.m. and 8:50 a.m. Based on past experience the company has determined that the employee is equally likely to arrive at any time between 8:00 a.m. and 8:50 a.m. Find the probability that the employee will arrive between 8:05 a.m. and 8:45 a.m

850-800= 50 845-805=40 40/50=.8 .8 is answer

a mean diameter of 140 millimeters, and a variance of 36. If a random sample of 50 steel bolts is selected, what is the probability that the sample mean would be less than 142.5 millimeters? Round your answer to four decimal places.

Normalcdf(0, 142.5, 140, 6/sqrt50)

Select all of the following indicators which would cause you to reject normality for a sample of data.

CHECK ALL OF THESE Having outliers Histogram of the data shows bar heights decreasing from left to right Normal probability of the data has no pattern UNCHECKED-the plot of the data follows linear pattern

A soft drink machine outputs a mean of 28 ounces per cup. The machine's output is normally distributed with a standard deviation of 3 ounces. What is the probability of filling a cup between 25 and 34 ounces? Round your answer to four decimal places

Normalcdf (25, 34, 28, 3)

The annual increase in height of cedar trees is believed to be distributed uniformly between five and twelve inches. Find the probability that a randomly selected cedar tree will grow less than 6 inches in a given year

area= Width x height height= 1/12-5= 1/7 width= 6-5=1 area= 1(1/7)

The Arc Electronic Company had an income of 72 million dollars last year. Suppose the mean income of firms in the same industry as Arc for a year is 60 million dollars with a standard deviation of 7 million dollars. If incomes for this industry are distributed normally, what is the probability that a randomly selected firm will earn more than Arc did last year? Round your answer to four decimal places.

normalcdf (72, 1000, 60, 7)

Consider the probability that greater than 91 out of 146 software users will not call technical support. Assume the probability that a given software user will not call technical support is 63%.

use binomial program first. 146 trials, probability .63. lower-91.5, upper-146. get standard and mean. Normalcdf (91.5, 146, 91.98, 5.8337)