We operate a bottle filling factory. One of our machines fills a 32-ounce bottle with a target of 32.08 ounces of orange juice. Every 10 minutes or so we obtain a sample filled bottle and hold it for the quality control department who will weigh the approximately 50 such bottles to determine if the machine is performing correctly. If the average of the 50 bottles is less than 32 ounces, the machine is shut down and the entire production from that 8-hour shift is held until further tests are performed.

a. We know that over the years this machine delivers 32.082 ounce of orange juice with a standard deviation 0.01 ounces. What is the probability that a sample of 50 bottles has a mean volume of less than 32 ounces?

b. There are 1,008 8-hour shifts per year (allowing for holidays, maintenance and a two-week scheduled shutdown). In a typical year, hour many times do we expect the quality department to shut down the machine and hold that shift’s production?

a. The probability that a sample of 50 bottles will have a mean volume less than 32 ounces is 0.0000 .

b. The number of times  we expect the quality department to shut down the machine and hold that shift’s production = 0.

Step-by-step explanation:

a. Letting X be the volume of orange juice in a bottle with u = 32.082,0 = 0.01

The distribution of X was not given,

Let \bar{X} be the sample mean of 50 bottles

Using the Central Limit Theorem, the sampling distribution of the sample mean \bar{X} follow Normal with

mean = 32.080 ( population mean )

Standard error = o/Vn=0.01/V50 = 0.0014

as the sample size is large

that is, X~ N(32.082, 0.00142

then _X - 32.082 ~ N(0.1) 0.0014

Finding the value of P(X <32)

P(X <32)

= P:<- 32 - 32.082 0.0014

= P(z < -58.57)

P(X <32) = 0.0000 (from z table)

The probability that a sample of 50 bottles will have a mean volume of fewer than 32 ounces is 0.0000.

b. The number of times we expect the quality department to shut down the machine

= 1008*P(X <32)

= 1008*(0)

= 0

Thus, the number of times we expect the quality department to shut down the machine and hold that shift’s production = 0.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this