Automobile Workers A worker in the automobile industry works an average of 43.1 hours per week. If the distribution isapproximately normal with a standard deviation of 1.5 hours, what is the probability that a randomly selected automobile workerworks less than 40 hours per week? Round the final answer to at least four decimal places and intermediate value calculations totwo decimal places.P(X<40) =

The Z-score for the random variable X equals 40 is obtained below as follows: Let X denotes the…

Answered: Automobile Workers A worker in the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Rochester, New York, averagesµ = 21.9 inches of snow for the month of December. The distribution of snowfall amounts is approximately normal with a standard deviation of σ =6.5 This year, a local jewelry store is advertising a refund of 50% off of all purchases made in December, if Rochester finishes the month with more than 3 feet (36 inches) of total snowfall. What is the probability that the jewelry store will have to pay off on its promise?
The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.885 grams and a standard deviation of 0.319 grams. Find the probability of randomly selecting a cigarette with 0.407 grams of nicotine or less. Round your answer to four decimals. P(X < 0.407) =
The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.974 grams and a standard deviation of 0.308 grams. Find the probability of randomly selecting a cigarette with 0.697 grams of nicotine or less. Round your answer to four decimals. P(X<0.697)=
Accordi the historical data, the life expectancy in Mexico is equal to the life expectancy in Spain. A new study has been made to see whether this has changea. Records of 205 individuals from Mexico who died recently are selected at random. The 205 individuals lived an average of 71.8 years with a standard deviation of 2.6 years. Records of 250 individuals from Spain who died recently are selected at random and independently. The 250 individuals lived an average of 71.2 years with a standard deviation of 3.0 years. Assume that the population standard deviation of the life expectancy can be estimated by the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the life expectancy, H1, in Mexico is not equal to the life expectancy, µ,, in Spain anymore? Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to at least three…
subpart. plz help with the last question
Private nonprofit four-year colleges charge, on average, $27,867 per year in tuition and fees. The standard deviation is $6,703. Assume the distribution is normal. Let X be the cost for a randomly selected college. Round all answers to 4 decimal places where possible. Find the probability that a randomly selected Private nonprofit four-year college will cost less than 31,421 per year. (___)Find the 62nd percentile for this distribution. $(____) (Round to the nearest dollar.)
K Assume that an adult female is randomly selected. Suppose females have pulse rates that are normally distributed with a mean of 71.0 beats per minute and a standard deviation of 11.7 beats per minute. Find Pg0, which is the pulse rate separating the bottom 90% from the top 10%. (Hint: Draw a graph.) Pg0 = (Round to one decimal place as needed.)
Americans on average drink 78 gallons of coffee individually each year, with a standard deviation of 13 gallons. Assume the population of amounts of coffee drank by Americans individually is normally distributed. If a sample of 15 Americans is taken, what is the probability that the sample mean amount of coffee drank individually will be more than 80 gallons? ? ( x̅ > 80 ??????? ) Round your final answer to four decimal places. = ______________
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 175.2-cm and a standard deviation of 1.8-cm. For shipment, 20 steel rods are bundled together.Find the probability that the average length of a randomly selected bundle of steel rods is greater than 176.2-cm.P(M > 176.2-cm) = ________Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Student scores on Professor of Stats final exam are normally distributed with a mean of 73 and a standard deviation of 6.5Find the probability of the following:**(use 4 decimal places)**a.) The probability that one student chosen at random scores above an 78. ________b.) The probability that 20 students chosen at random have a mean score above an 78. _________c.) The probability that one student chosen at random scores between a 68 and an 78. __________d.) The probability that 20 students chosen at random have a mean score between a 68 and an 78. __________
Lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 13 days. A baby is considered premature if the pregnancy length is at or below the third percentile (P3) of all pregnancy lengths. Find the length of pregnancy in days that separates the premature babies from those that are not considered premature. Round to the nearest integer.

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