PROBLEM SOLVING Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during
1 answer:
Response:
The likelihood that no more than 50 seals were spotted during a randomly selected survey is 0.0516.
Step-by-step breakdown:
It is stated that Scientists performed aerial surveys of a seal sanctuary and recorded the variable x indicating the number of seals observed in each survey.
The counts of seals observed followed a normal distribution characterized by a mean of 73 seals and a standard deviation of 14.1 seals.
Let X = the count of seals observed
The distribution for the z-score in a normal population is defined as;
Z = ~ N(0,1)
where, = average number of seals observed = 73
= standard deviation = 14.1
Now, to find the probability that at most 50 seals were spotted during a randomly selected survey, we compute = P(X 50 seals)
P(X 50) = P(
) = P(Z
-1.63) = 1 - P(Z < 1.63)
= 1 - 0.94845 = 0.0516
The above probability was calculated using the z-table for the value of x = 1.63 which corresponds to an area of 0.94845.
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