Real populations sometimes increase beyond their environment’s carrying capacity in a relatively short period of time. What is t
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Answer:
P(-1 < z < 1) = 0.3174
Step-by-step explanation:
Mean (μ) = 1.62 ounces
Standard Deviation (σ) = 0.05
No of balls (sample size n) = 100
X = weight of a ball
Weight of a group of 100 balls must lie in the range 162 ± 0.5 ounces i.e. weight of a single ball will be 162/100 ± 0.5/100 ounces = 1.62 ± 0.005 ounces.
So, we need to find the probability P (1.615 < X < 1.625). We will use the central limit theorem.
z = (Χ' - μ)/(σ/)
P (1.615 < X < 1.625) = ( < (Χ - μ)/(σ/
) <
)
= (-1 < z < 1)
We need to find the probability of P (-1 < z < 1) by looking at the Normal Distribution Probability Table.
In order to make our working simpler, we need to break P (-1 < z < 1) into two parts: P(z < 1) and P(z > -1)
The probability for areas under the normal curve are given for P(z>X) so we can directly find the probability of P (z > -1) by referring to the normal probability table.
P(z > -1) = 0.1587
We can calculate P(z < 1) by subtracting P(z >1) from the total probability (i.e. 1). P(z >1) can be obtained from the normal probability table.
P(z < 1) = 1 - 0.8413 = 0.1587
By adding the two probabilities together, we get:
P(-1 < z < 1) = P(z < 1) + P (z > -1)
= 0.1587 + 0.1587
P(-1 < z < 1) = 0.3174