Find the value of x and y.
2 answers:
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Answer:
Probability that a randomly selected pregnancy lasts less than 233 days is 0.3594.
Step-by-step explanation:
We are given that the lengths of the pregnancies of a certain animal are approximately normally distributed with mean mu equals 238 days and standard deviation sigma equals 14 days.
Let X = <u><em>lengths of the pregnancies of a certain animal</em></u>
So, X ~ Normal()
The z score probability distribution for normal distribution is given by;
Z = ~ N(0,1)
where, = population mean = 238 days
= standard deviation = 14 days
Now, the probability that a randomly selected pregnancy lasts less than 233 days is given by = P(X < 233 days)
P(X < 233 days) = P( <
) = P(Z < -0.36) = 1 - P(Z
0.36)
= 1 - 0.6406 = 0.3594
The above probability is calculated by looking at the value of x = 0.36 in the z table which has an area of 0.6406.