Answer:
Probability that at most 50 seals were observed during a randomly chosen survey is 0.0516.
Step-by-step explanation:
We are given that Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during each survey.
The numbers of seals observed were normally distributed with a mean of 73 seals and a standard deviation of 14.1 seals.
Let X = <u><em>numbers of seals observed</em></u>
The z score probability distribution for normal distribution is given by;
Z = 
~ N(0,1)
where, 
= population mean numbers of seals = 73
= standard deviation = 14.1
Now, the probability that at most 50 seals were observed during a randomly chosen survey is given by = P(X 
50 seals)
P(X 50) = P( 
) = P(Z -1.63) = 1 - P(Z < 1.63)
= 1 - 0.94845 = <u>0.0516</u>
The above probability is calculated by looking at the value of x = 1.63 in the z table which has an area of 0.94845.
<span>When you are sampling from a small finite lot, the hypergeometric distribution applies. The binomial is a poor approximation in this case.
The general equation for the hypergeometric where aCx means the number of combinations of a items selected x at-a-time.:
P(x) =[(aCx)(N-aCn-x)]/NCn
Where
N is the lot size = 20
a is the number of defectives in the lot = 3.
x is the number of defectives in the sample.
n is the sample size = 2.
A. The probability that the first item is defective is
P(x=1) = [(3C1)(17C1)]/(20C2)
= (3)(17)/190 = 0.268
The probability that the second item is defective is
P(x = 1) = [(2C1)(17C1)]/(19C2) = (2)(17)/171 = 0.199.
So the total probability is (0.268)(0.199) = 0.0532
B. The probability that the first item is good is:
P(x = 0) = (3C0)(16C2)]/20C2 = (1)(120)/190 = 0.632
The probability that the second item is defective is
P(x = 0) =[(3C0)(16C2)]/19C2
= (1)(120)/171 = 0.670.
The total probability is 0.632(0.670) = 0.4234</span>
In scientific notation, a number is usually?.?*10^??
for example:
1.3*10^-2
To write it in standard notation you can simply multiply it out
for example:
0.013
