Answer:
p(on schedule) ≈ 0.7755
Step-by-step explanation:
A suitable probability calculator can show you this answer.
_____
The z-values corresponding to the build time limits are ...
z = (37.5 -45)/6.75 ≈ -1.1111
z = (54 -45)/6.75 ≈ 1.3333
You can look these up in a suitable CDF table and find the difference between the values you find. That will be about ...
0.90879 -0.13326 = 0.77553
The probability assembly will stay on schedule is about 78%.
Answer:
Step-by-step explanation:
Step one:
given data
heigth/depth = 3.5m
radius= 9m
Step two:
The shape is that of a cylinder
Hence the volume of the pool is
substitute
Complete Question
The complete question is shown on the first uploaded image
Answer:
The probability that there exist 60 or more defected children is
Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence
Step-by-step explanation:
From the question we are told that
in every 1000 children a particular genetic defect occurs to 1
The number of sample selected is
The probability of observing the defect is mathematically evaluated as
The probability of not observing the defect is mathematically evaluated as
The mean of this probability is mathematically represented as
Substituting values
The standard deviation of this probability is mathematically represented as
Substituting values
the probability of detecting defects can be represented in as normal distribution like
in standardizing the normal distribution the normal area used to approximate is the right of 59.5 instead of 60 because x= 60 is part of the observation
The z -score is obtained mathematically as
The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve
Note: We are looking for the area to the right i.e 60 or more
The total area under the curve is 1
So
