Complete question is:
Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per pound, with a standard deviation of 7 grams of fat per pound. A random sample of 34 farm-raised trout is selected. The mean fat content for the sample is 29.7 grams per pound. Find the probability of observing a sample mean of 29.7 grams of fat per pound or less in a random sample of 34 farm-raised trout. Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.
Answer:
Probability = 0.0277
Step-by-step explanation:
We are given;
Mean: μ = 32
Standard deviation;σ = 7
Random sample number; n = 34
To solve this question, we would use the equation z = (x - μ)/(σ/√n) to find the z value that corresponds to 29.7 grams of fat.
Thus;
z = (29.7 - 32)/(7/√34)
Thus, z = -2.3/1.200490096
z = -1.9159
From the standard z table and confirming with z-calculator, the probability is 0.0277
Thus, the probability to select 34 fish whose average grams of fat per pound is less than 29.7 = 0.0277
Answer:
Add
Step-by-step explanation:
10.65*40=426
426+50=476
Answer: 8
Step-by-step explanation:
Answer: 1.95
Step-by-step explanation:
Multiply 15*13%
Multiply 5 and £14.06 to get £70.30 this is your answer
