## Normal distribution

1. X is a normally distributed variable with mean μ = 50 and standard deviation σ = 8. Find

• a) P(x < 60)
• b) P(x >40)
• c) P(45 < x < 65)

Solution:
• a) For x = 60,
• z = (60 - 50) / 8 = 1.5
• Hence P(x < 60) = P(z < 1.5) = [area to the left of 1.5] = 0.8944

• b) For x = 40, z = (40 - 50) / 8 = -1.25
• Hence P(x > 40) = P(z > -1.25) = [total area] - [area to the left of -1.25]
• = 1 - .1056= 0.8944

• c) For x = 45 , z = (45 - 50) / 8 = 0.63 and for x = 65, z = (65 - 50) / 8 = 1.88
• Hence P(45 < x < 65) = P(0.63 < z < 1.88) = [area to the left of z = 1.88] - [area to the left of 0.63]
• = 0.8944 - 0.2659 = 0.6284

2. In F-1 race the speeds are normally distributed with a mean of 75 km/hr and a standard deviation of 5 km/hr. What is the probability that a racing car picked at random is travelling at more than 80 km/hr?

3. Solution:
• Let x be the random variable that represents the speed of racing cars as x has μ = 75 and σ = 5. We have to find the probability that x is higher than 100 or P(x > 100)
• For x = 80 , z = (80 - 75) / 5 = 1
• P(x > 80) = P(z > 1) = [total area] - [area to the left of z = 1]
• = 1 - 0.8413 = 0.1587
• The probability that a racing car selected at a random has a speed greater than 80 km/hr is equal to 0.1587

4. The length of time between the completion of a particular task is normally distributed with a mean of 50 hours and a standard deviation of 12 hours. Ron finishes those task and wants to know the probability that the length of time will be between 52 and 68 hours.
5. Solution:
• Let x 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 70.
• For x = 50, z = (50 - 50) / 15 = 0
• For x = 70, z = (70 - 50) / 15 = 1.33
• P (50< x < 70) = P (0< z < 1.33) = [area to the left of z = 1.33] - [area to the left of z = 0]
• = 0.9082 - 0.5 = 0.4082
• The probability that John's computer has a length of time between 50 and 70 hours is equal to 0.4082.

6. Academic assessment for entering into defence field is determined through scores. This tests are normally distributed with a mean of 650 and a standard deviation of 150. Shaun wants to be in defence and he knows that he must score better than at least 70% of the aspirants who took the test. Shaun takes the test and scores 700. Will he get through in defence?
7. Solution:
• Let x be the random variable that represents the scores. x is normally distributed with a mean of 750 and a standard deviation of 150. 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 by 100, we obtain percentages.
• For x = 650, z = (700 - 650) / 150 = 0.33
• For x = 650, z = (700 - 650) / 150 = 0.33
• The proportion P of students who scored below 700 is given by
• P = [area to the left of z = 0.33] = 0.63 = 63%
• Shaun scored better than 63% of the students who took the test and he will not be admitted to defence.

8. The length of potatoes produced are approximated by a normal distribution model with a mean of 5 cm and a standard deviation of 0.2 cm. If a potato is chosen at random
9. a) what is the probability that the length of this component is between 4.80 cm and 5.20 cm?

b) what is the probability that the length of this component is between 4.94 and 5.6 cm?

Solution:
• P (4.80 < x < 5.20) = P (-1 < z < 1)= 0.6826b) P (4.94 < x < 5.4) = P (-3 < z < 3)= 0.9772 – 0.3820
• = 0.5951

10. The life of a frog has a normal distribution with a mean of 12 months and standard deviation of 2 months. Find the probability that a randomly selected frog will last.
11. a) less than 7 months.

b) between 7 and 12 months.

Solution:
• P(x < 7) = P(z < -2.5)= 0.0062
• P(7 < x < 12) = P(-2.5 < z < 0)= 0.4938

12. The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 556 MPa with a standard deviation of 10 MPa. (a) What is the probability that a randomly chosen sample of glass will break at less than 556 MPa? (Round your answer to 2 decimal places.) Probability (b) What is the probability that a randomly chosen sample of glass will break at more than 575 Mpa? (Round your answer to 4 decimal places.) Probability (c) What is the probability that a randomly chosen sample of glass will break at less than 588 MPa? (Round your answer to 4 decimal places.) Probability
13. mean = 556

std = 10

Solution:
• a) for x =556
• z= (556-556)/10 = 0
• P(x< 581) = P(z< 0)= 0.5000,
• b) for x = 575
• z = (575 - 556)/10 = 19/10
• =1.9
• P(x > 575) = P(z > 1.9) =0.0287

14. Assume that body temperatures of healthy adults are normally distributed with a mean of 98.2°F and a standard deviation of 0.62°F. Suppose we take a sample of eight healthy adults. What is the probability that their mean body temperature will be greater than 98.4?
15. Solution:
• mean = 98.2
• standard deviation = 0.62
• n=8
• X= 98.4
• p(X>98.4)
• z = (98.4 - 98.2) / 0.62/sqrt(8)
• = 0.2 / 0.219
• = 0.91
• p(X<98.4) = 0.8186
• p(X>98.4) = 1 - 0.8186
• = 0.1814
16. According to the US Bureau of Labor Statistics estimates, the average earnings of construction workers were \$18.96 per hour in August 2002.Assume that current earnings of all construction workers are normally distributed with a mean of \$18.96 per hour and a standard deviation of \$3.60 per hour. Find the probability that the mean hourly earnings of a random sample of 25 construction workers is:
A) Betwwn \$18 and \$20 an hour ?
B) Greater than the population mean by \$1.50 of more?

17. Solution:

xbar = 18.96
std = 3.60

• A)  For x=18
z = (x - xbar) / std
= ( 18 - 18.96) / 3.60
= -0.96 / 3.60
= -0.266
z = (20 - 18.96 ) / 3.60
= 0.26
p( -0.27 < z < 0.26 ) = 0.6026 - 0.3974
= 0.2050

• B)  x = 18.96+1.50 = 20.46
Z = ( 20.46 - 18.96) / 3.60
= 0.42
p( x>=20.46) = 0.6628

18. A manufacturer of sports equipment claims that its synthetic fishing line has a mean breaking strength of 8 kilograms. A random sample of 50 lines is tested and found to have a mean breaking strength of 7.8 kilograms. Does this data indicate that the population mean breaking strength is different than(either higher or lower than) 8 kilograms?Assume we ate sampling from a normal population with standard deviation σ0.5. Use a 0.01 level of significance.

A) State the null hypotesis and the alternative hypothesis.
B) Identify the rejection region.
C) Compute the test statistic.
D) Draw your conclusion. Is there sufficient evidence to conclude that the mean is different than 8?
E) What is the p-value of the test statistic?

19. Solution:

• A) Null Hypothesis : The population mean breaking is equal to 8.
Alternative Hypothesis: The population mean breaking is different than 8.

• B) Rejection region from table
The rejection level at level of significance 0.01 is 2.58

• C) Test Statistic:
SE = 0.5 / sqrt(50) = 0.07
Z = (7.8 - 8) / SE
= (7.8 - 8) / 0.07
= -2.85

• D) Since the calculated value -2.85 falls beyond theinterval ( -2.57 , 2.57), hence we reject null hypothesis and conclude that their is a sufficient evidence to conclude that mean is differnt than 8.

• E) P-value = 0.9978

20. a normal population has mean of 62 and a standard deviation of 12. you select a random sample of 16. compute probability the sample mean.
• a) greater than 66
b) less than 59
c) between 59 and 66

Solution:
• a)
Given,
mu     =  62
X    =  66
n     =  16
Standard Deviation (SD) =   12
Standard Error (S.E)    =    3
Z = ( X - mu) / SE
Z = 1.33
p( x> 66) = p(z>1.33) = 0.09176

• b)
Given,
mu     =  62
X    =  66
n     =  16
Standard Deviation (SD) =   12
Standard Error (S.E)    =    3
Z = ( X - mu) / SE
Z = -1
p(x<59) = p(z< -1) = 0.15866

• c)
p( 66 < x <59) = p( 1.33 < Z < -1)
= 0.90824 - 0.15866 = 0.7496