Answer:
Step-by-step explanation:
2081.25 ; 2312.50 ; 2543.75 ; ..... 7500
a= 2081.25
d = 231.25
2081.25 + (n-1) * 231.25 = 7500
(n -1) *231.25 = 7500 - 2081.25
231.25n - 231.25 = 5418.75
231.25n = 5418.75 + 231.25
231.25n = 5650
n = 5650/231.25 = 565000/23125
n = 904/37
n = 24 years 157 days
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
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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