A fertilizer company manufactures 10-pound bags of fertilizer with a standard deviation of 0.24 pounds per bag. The bag weights are normally distributed. What is the probability that a sample of 4 bags will have a mean weight less than 9.8 pounds

the probability that a sample of 4 bags will have a mean weight less than 9.8 pounds is 0.05

Step-by-step explanation:

Given the data in the question;

μ_x = 10 pound bags

standard deviation s_x = 0.24 pounds

sample size n = 4

The bag weights are normally distributed so;

p( x' less than 9.8 ) will be;

p(  (x'-μ_x' / s_x')   <   (9.8-μ_x' / s_x')  )

we know that;

μ_x' = μ_x = 10

and s_x' = s_x/√n = 0.24/√4

so; we substitute

p(  z  <  ( (9.8 - 10) / (0.24/√4)  )

p(  z  <  -0.2 / 0.12   )

p(  z  <  -1.67   )

{ From z-table }

⇒ p(  z  <  -1.67   ) = 0.0475 ≈ 0.05

Therefore, the probability that a sample of 4 bags will have a mean weight less than 9.8 pounds is 0.05

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1