Respuesta :
Answer:
95.02% of squirrels have a weight between 3 and 5 pounds.
Step-by-step explanation:
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 4.1 pounds
Standard Deviation, σ = 0.5 pounds
We are given that the distribution of weight is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(weight between 3 and 5 pounds)
[tex]P(3 \leq x \leq 5) = P(\displaystyle\frac{3 - 4.1}{0.5} \leq z \leq \displaystyle\frac{5-4.1}{0.5}) = P(-2.2 \leq z \leq 1.8)\\\\= P(z \leq 1.8) - P(z < -2.2)\\= 0.9641- 0.0139 = 0.9502= 95.02\%[/tex]
[tex]P(3 \leq x \leq 5) = 95.02\%[/tex]
95.02% of squirrels have a weight between 3 and 5 pounds.
The percentage of squirrels that have a weight between 3 and 5 pounds is; 95.017%
We are given;
Mean; μ = 4.1 pounds
Standard Deviation; σ = 0.5 pounds
Since this is a normal distribution, then formula for the test statistic which is the z-score is;
z = (x' - μ)/σ
We want to find the percentage of squirrels have a weight between 3 and 5 pounds. This means;
P(3 ≤ x' ≤ 5)
Let us first find z-score at x' = 3;
z = (3 - 4.1)/0.5
z = -1.1/0.5
z = -2.2
Let us do the same at x' = 5;
z = (5 - 4.1)/0.5
z = 0.9/0.5
z = 1.8
From online calculator of probability between two z-scores attached, we have;
P(-2.2<x<1.8) = 0.95017
Thus; P(-2.2<x<1.8) = 95.017 %
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