Question

1. If Z is a standard normal random variable, find the value z₀ for the following probabilities. (Round your answers to two decimal places.)

(a) P(Z > z₀) = 0.5

z₀ =

(b) P(Z < z₀) = 0.8686

z₀ =

(c) P(−z₀ < Z < z₀) = 0.90

z₀ =

(d) P(−z₀ < Z < z₀) = 0.99

z₀ =

2. A company that manufactures and bottles apple juice uses a machine that automatically fills 64-ounce bottles. There is some variation, however, in the amount of liquid dispensed into the bottles. The amount dispensed has been observed to be approximately normally distributed with mean 64 ounces and standard deviation 1 ounce.

Determine the proportion of bottles that will have more than 65 ounces dispensed into them. (Round your answer to four decimal places.)

________ oz

3. The grade point averages (GPAs) of a large population of college students are approximately normally distributed with mean 2.5 and standard deviation 0.8.

If students possessing a GPA less than 1.7 are dropped from college, what percentage of the students will be dropped? (Round your answer to two decimal places.)

_______ %

4. A soft-drink machine can be regulated so that it discharges an average of μ ounces per cup. If the ounces of fill are normally distributed with standard deviation 0.7 ounce, give the setting for μ so that 8-ounce cups will overflow only 3% of the time. (Round your answer to three decimal places.)

μ = _______ oz

Answer 1

Solution :

Given that

1) Using standard normal table,

a)  P(Z > z₀) = 0.5

= 1 -  P(Z < z₀) = 0.5

= P(Z < z₀) = 1 - 0.5

= P(Z < z₀) = 0.5

= P(Z < 0) = 0.5

z₀ = 0

b) P(Z < z₀) = 0.8686

= P(Z < 1.12) = 0.8686

= z₀ = 1.12

c) P( - z₀ < Z < z₀) = 0.90

= P(Z < z₀) - P(Z <- z₀ ) = 0.90

= 2P(Z < z₀) - 1 = 0.90

= 2P(Z < z₀) = 1 + 0.90

= P(Z < z₀) = 1.90 / 2

= P(Z < z₀) = 0.95

= P(Z < 1.65) = 0.95

= z₀ ± 1.65

d) P( - z₀ < Z < z₀) = 0.99

= P(Z < z₀) - P(Z <- z₀ ) = 0.99

= 2P(Z < z₀) - 1 = 0.99

= 2P(Z < z₀) = 1 + 0.99

= P(Z < z₀) = 1.99 / 2

= P(Z < z₀) = 0.995

= P(Z < 2.58) = 0.995

= z₀ ± 2.58

2) mean = = 64

standard deviation = = 1

P(x > 65) = 1 - p( x< 65)

=1- p P[(x - ) / < (65 - 64) / 1]

=1- P(z <1.00 )

Using z table,

= 1 - 0.8413

= 0.1587

3) mean = = 2.5

standard deviation = = 0.8

P(x < 1.7) = P[(x - ) / < (1.7 - 2.5) / 0.8]

= P(z < -1.00)

Using z table,

= 0.1587

The percentage is = 15.87%

4) mean = = 8

standard deviation = = 0.7

Using standard normal table,

P(Z > z) = 3%

= 1 - P(Z < z) = 0.03  

= P(Z < z) = 1 - 0.03

= P(Z < z ) = 0.97

= P(Z < 1.88 ) = 0.97  

z = 1.88

Using z-score formula,

x = z * +

x = 1.88 * 0.7 + 8

x = 9.316 oz