What Percent Of A Standard Normal Model Is Found In Each Region? Be Sure To Draw A Picture First.

Normal Distribution Issues with Solutions

Problems and applications on normal distributions are presented. The solutions to these bug are at the lesser of the page. An online normal probability estimator and an inverse normal probability calculator may be useful to bank check your answers.

Issues

X is a normally usually distributed variable with mean μ = 30 and standard departure σ = 4. Find
a) P(x < 40)
b) P(x > 21)
c) P(thirty < 10 < 35)

A radar unit of measurement is used to measure speeds of cars on a thruway. The speeds are normally distributed with a mean of 90 km/hr and a standard deviation of ten km/hour. What is the probability that a car picked at random is travelling at more than than 100 km/hour?

For a certain type of computers, the length of fourth dimension bewteen charges of the battery is normally distributed with a mean of 50 hours and a standard deviation of 15 hours. John owns one of these computers and wants to know the probability that the length of time will exist between 50 and 70 hours.
Entry to a certain Academy is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Tom wants to be admitted to this university and he knows that he must score better than at least seventy% of the students who took the test. Tom takes the test and scores 585. Will he be admitted to this academy?

The length of similar components produced by a visitor are approximated by a normal distribution model with a hateful of 5 cm and a standard deviation of 0.02 cm. If a component is chosen at random
a) what is the probability that the length of this component is between iv.98 and 5.02 cm?
b) what is the probability that the length of this component is between 4.96 and 5.04 cm?
The length of life of an instrument produced past a machine has a normal ditribution with a mean of 12 months and standard deviation of 2 months. Find the probability that an instrument produced by this motorcar volition final
a) less than 7 months.
b) betwixt seven and 12 months.
The time taken to assemble a car in a certain plant is a random variable having a normal distribution of 20 hours and a standard difference of ii hours. What is the probability that a car tin be assembled at this plant in a period of time
a) less than 19.five hours?
b) betwixt 20 and 22 hours?
A large group of students took a examination in Physics and the final grades have a mean of lxx and a standard deviation of 10. If we can approximate the distribution of these grades by a normal distribution, what percent of the students
a) scored higher than 80?
b) should pass the test (grades≥sixty)?
c) should fail the test (grades<60)?
The almanac salaries of employees in a large company are approximately normally distributed with a mean of $50,000 and a standard deviation of $twenty,000.
a) What percent of people earn less than $twoscore,000?
b) What percent of people earn between $45,000 and $65,000?
c) What percent of people earn more than than $lxx,000?

Solutions to the In a higher place Problems

Notation: What is meant here by surface area is the area under the standard normal curve.
a) For x = 40, the z-value z = (twoscore - 30) / 4 = 2.5
Hence P(ten < xl) = P(z < ii.5) = [area to the left of two.5] = 0.9938
b) For 10 = 21, z = (21 - 30) / four = -two.25
Hence P(x > 21) = P(z > -two.25) = [total area] - [area to the left of -2.25]
= ane - 0.0122 = 0.9878
c) For ten = 30 , z = (30 - thirty) / four = 0 and for x = 35, z = (35 - 30) / four = 1.25
Hence P(30 < x < 35) = P(0 < z < i.25) = [area to the left of z = 1.25] - [surface area to the left of 0]
= 0.8944 - 0.v = 0.3944
Let 10 be the random variable that represents the speed of cars. x has μ = 90 and σ = x. We have to find the probability that x is higher than 100 or P(x > 100)
For x = 100 , z = (100 - 90) / 10 = 1
P(x > xc) = P(z > 1) = [full area] - [area to the left of z = 1]
= ane - 0.8413 = 0.1587
The probability that a car selected at a random has a speed greater than 100 km/hr is equal to 0.1587
Let ten be the random variable that represents the length of time. It has a mean of 50 and a standard deviation of 15. We have to find the probability that x is between 50 and lxx or P( fifty< 10 < 70)
For ten = 50 , z = (50 - 50) / 15 = 0
For x = 70 , z = (70 - 50) / fifteen = i.33 (rounded to ii decimal places)
P( 50< x < 70) = P( 0< z < one.33) = [area to the left of z = one.33] - [area to the left of z = 0]
= 0.9082 - 0.five = 0.4082
The probability that John's computer has a length of time betwixt l and 70 hours is equal to 0.4082.
Allow 10 be the random variable that represents the scores. 10 is unremarkably ditsributed with a mean of 500 and a standard deviation of 100. The total area under the normal curve represents the total number of students who took the test. If we multiply the values of the areas under the curve past 100, we obtain percentages.
For ten = 585 , z = (585 - 500) / 100 = 0.85
The proportion P of students who scored below 585 is given by
P = [area to the left of z = 0.85] = 0.8023 = 80.23%
Tom scored better than 80.23% of the students who took the test and he will be admitted to this University.
a) P(4.98 < x < v.02) = P(-one < z < i)
= 0.6826
b) P(iv.96 < x < 5.04) = P(-2 < z < ii)
= 0.9544
a) P(x < vii) = P(z < -two.v)
= 0.0062
b) P(7 < x < 12) = P(-2.5 < z < 0)
= 0.4938
a) P(10 < 19.5) = P(z < -0.25)
= 0.4013
b) P(xx < 10 < 22) = P(0 < z < 1)
= 0.3413
a) For 10 = eighty, z = 1
Surface area to the correct (higher than) z = 1 is equal to 0.1586 = 15.87% scored more that 80.
b) For ten = 60, z = -1
Area to the right of z = -ane is equal to 0.8413 = 84.13% should pass the test.
c)100% - 84.xiii% = 15.87% should fail the test.
a) For x = 40000, z = -0.v
Surface area to the left (less than) of z = -0.5 is equal to 0.3085 = 30.85% earn less than $twoscore,000.
b) For x = 45000 , z = -0.25 and for x = 65000, z = 0.75
Expanse between z = -0.25 and z = 0.75 is equal to 0.3720 = 37.20 earn between $45,000 and $65,000.
c)For x = 70000, z = i
Area to the right (higher) of z = 1 is equal to 0.1586 = xv.86% earn more than $lxx,000.