Question

3. In recent years, the results of a particular college entrance exam showed an average score of 55 with a standard deviation of 4 points.

a. Based on these results, what is the probability that an incoming student scores below a 50?

b. What is the probability that a student scores above a 60?

c. What is the probability that a student scores between a 53 and a 59?

d. What is the probability that a student scores below a 58?

4. A bag of potato chips claims to contain 7 oz. of potato chips. Random sampling determines that the bags contain an average of 7.2 oz. with a standard deviation of 0.081 oz.

a. What is the maximum weight that 20% of the bags contain less than?

b. What is the minimum weight that the heaviest 15% of the bags contain?

c. In what weight interval are the middle 60% of the bags contained?

Answer 1

Solution :

Given that ,

3)mean = = 55

standard deviation = = 4

a) P(x < 50) = P[(x - ) / < (50 -55) /4 ]

= P(z < -1.25 )

= 0.1056

probability =0.1056

b)

P(x > 60) = 1 - p( x< 60)

=1- p [(x - ) / < (60 -55) /4 ]

=1- P(z < 1.25)

= 1 - 0.8944 = 0.1056

probability =0.1056

c)

P(53< x < 59) = P[(53-55)/4 ) < (x - ) /  < (59-55) /4 ) ]

= P(-0.50 < z < 1)

= P(z <1 ) - P(z <-0.50 )

Using standard normal table

= 0.8413 - 0.3085 = 0.5328

Probability =0.5328

d)

P(x < 58) = P[(x - ) / < (58-55) /4 ]

= P(z < 0.75 )

= 0.7734

probability =0.7734

4)

mean = = 7.2

standard deviation = = 0.081

a) P(Z < z ) = 0.20

z =-0.84

Using z-score formula,

x = z * +

x = -0.84 * 0.081+7.2

x = 7.13

Maximum value = 7.13

b)

P(Z > z ) = 0.15

1- P(z < z) =0.15

P(z < z) = 1-0.15 = 0.85

z =1.036

Using z-score formula,

x = z * +

x = 1.036*0.081+7.2

x = 7.28

Minimum value = 7.28

c) middle 60 %

P(-z Z z) = 0.60

P(Z z) - P(Z -z) = 0.60

2P(Z z) - 1 = 0.60

2P(Z z) = 1 + 0.60= 1.60

P(Z z) = 1.60 / 2 = 0.80

P(Z 0.84) = 0.80

z = 0.84

z = -0.84

x = z * +

x = -0.84 * 0.081+7.2

x = 7.13

z = +0.84.

x = z * +

x = 0.84 * 0.081+7.2

x = 7.27

interval weight = 7.13 and 7.27