Please help me get this done. due at 11:59 tonight
1 answer:
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The probability that a pregnancy last at least 300 days is 0.01659
Step-by-step explanation:
The formula of z-score is z = (x - μ)/σ, where
- μ is the mean
- σ is the standard deviation
- x is the score
The lengths of human pregnancies are normally distributed with a
mean of 268 day & a standard deviation of 15 days
∴ μ = 268 days
∴ σ = 15 days
We need to find the probability that a pregnancy last at least 300 days
∵ At least means greater than or equal
∴ x ≥ 300 days
For probability that x ≥ 300 find the z-score and use the normal
distribution table to find the area to the right of the z-score
∵ z = (x - μ)/σ
∴
∴ <em>z</em> = 2.13
By using the normal distribution table of z-score
∵ The area (to the left of z-score) corresponding to z-score of 2.13
is 0.98341
∵ We need the area to the right of z-score
∴ P(x ≥ 300) = 1 - 0.98341
∴ P(x ≥ 300) = 0.01659
The probability that a pregnancy last at least 300 days is 0.01659
Learn more:
You can learn more about probability in brainly.com/question/9178881
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