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Adult racehorse weights are normally distributed, with a mean of 500 kg and a standard deviation of 33 kg. Approximately what percentage of the adult racehorses weigh more than 570 kg?
1.7%
98.3%
93.3%
0.12%

2 Answers

5 votes

Answer:

yes it is correct good work

User Oakymax
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Answer:

The percentage of the adult racehorses weigh more than 570 kg is 1.7% ⇒ 1st answer Step-by-step explanation: Adult racehorse weights are normally distributed, with a mean of 500 kg and a standard deviation of 33 kg We need to find the percentage of the adult racehorses weigh more than 570 kg To solve the problem do these steps 1. Find z-score 2. Use z-score to find the corresponding area which represented P(x) 3. Change P(x) to percentage ∵ z = (x - μ)/σ ∵ The mean μ = 500 kg ∵ The standard deviation σ = 33 kg ∵ x = 570 kg Substitute these values in the rule above ∴ z = Let us use the normal distribution table to find the corresponding area ∵ The area corresponding to z-score of 2.12 is 0.98300 ∵ We need x > 570 , then we need the area to the right of the z-score ∴ P(x > 570) = 1 - 0.98300 = 0.017 Change it to percentage by multiplying the answer by 100% ∴ The percentage of x > 570 = 0.017 × 100% = 1.7% The percentage of the adult racehorses weigh more than 570 kg is 1.7%

Explanation:

The percentage of the adult racehorses weigh more than 570 kg is 1.7% ⇒ 1st answer Step-by-step explanation: Adult racehorse weights are normally distributed, with a mean of 500 kg and a standard deviation of 33 kg We need to find the percentage of the adult racehorses weigh more than 570 kg To solve the problem do these steps 1. Find z-score 2. Use z-score to find the corresponding area which represented P(x) 3. Change P(x) to percentage ∵ z = (x - μ)/σ ∵ The mean μ = 500 kg ∵ The standard deviation σ = 33 kg ∵ x = 570 kg Substitute these values in the rule above ∴ z = Let us use the normal distribution table to find the corresponding area ∵ The area corresponding to z-score of 2.12 is 0.98300 ∵ We need x > 570 , then we need the area to the right of the z-score ∴ P(x > 570) = 1 - 0.98300 = 0.017 Change it to percentage by multiplying the answer by 100% ∴ The percentage of x > 570 = 0.017 × 100% = 1.7% The percentage of the adult racehorses weigh more than 570 kg is 1.7%

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User Xi Sigma
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