Question

The weights of the fish in a certain lake are normally distributed with a
mean of 20 lb and a standard deviation of 9. If 9 fish are randomly selected,
what is the probability that the mean weight will be between 17.6 and 23.6
Ib? Write your answer as a decimal rounded to 4 places

Answer 1

$P(17.6< xbar < 23.6) = P(-0.8 < t < 1.2 when df = 3) = 0.6730$

Solution

Normal distribution between 17.6 and 23.6

μ  = 20 lb

σ/x  = σ /√n = 9/√9 = 3

Distribution score for  17.6 = 17.6 - 20÷ 3 = - 0.8

$Distribution score for 17.6 = 17.6 - 20÷ 3 = - 0.8$

Distribution score for 23.6 = 23.6 - 20÷ 3 = 1.2

$Distribution score for 23.6 = 23.6 - 20÷ 3 = 1.2$

p(mean weight between 17.6 and 23.6 for 9 fish) =

9(-0.8 ∠≠∠1.2) = 0.8849 - 0,2119 = 0.6730

P(17.6< xbar < 23.6) = P(-0.8 < t < 1.2 when df = 3) = 0.6730

$P(17.6< xbar < 23.6) = P(-0.8 < t < 1.2 when df = 3) = 0.6730$

Answer 2

The 0.6730 is the probability which is the mean weight of the given equation and the further calculation can be defined as follows:

Sampling distribution:

Mean $(\mu) = 20\ lb$

Standard deviation $(\sigma) = 9$

Sample size $(n) = 9$

Sampling distribution of average $\bar{x}$, per the Central Limit Theorem, equals normal.

$\to P(\bar{x} < A) = P(Z < (A - \frac{\mu_{\bar{x}})}{\sigma_{\bar{x}}})\\\\\to \mu_{\bar{x}} = \mu = 20\ lb\\\\\to \sigma_{\bar{x}} = \frac{\sigma }{\sqrt{n}}$

         $= \frac{9}{\sqrt{9}}\\\\= \frac{9}{3}\\\\= 3$

P(the mean weight will be between 17.6 and 23.6 lb):

$= P(17.6 < \bar{x} < 23.6)$

$= P(\bar{x} < 23.6) - P(\bar{x} < 17.6)\\\\= P(Z < \frac{(23.6 - 20)}{3}) - P(Z < \frac{(17.6 - 20)}{3})\\\\= P(Z < 1.2) - P(Z < -0.8)\\\\= 0.8849 - 0.2119\\\\= 0.6730$

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