Question

QUESTION 6 & 7.

The next three questions (5 - 7) refer to the following: Weights of walleye from the Lake Winnipeg follow a normal distribution with mean 11 kg and standard deviation 2.5 kg. Question 5 (1 point) ✓ Saved What is the probability that the average weight of a random sample of 5 walleye is less than 10 kg? Keep 4 decimal places in intermediate calculations and report your final answer to 4 decimal places. Your Answer: 0.1038 Answer Question 6 (1 point) What is the probability that the average weight of a random sample of 10 walleye is greater than 12 kg? Keep 4 decimal places in intermediate calculations and report your final answer to 4 decimal places. Your Answer:

Answer 1

Solution :

Let X be a random variable which represents the weights of walleye from the lake Winnipeg.

Given that, X ~ N(11, 2.5²)

μ = 11 kg and  σ = 2.5 kg

6) We have to find P(x̄ > 12).

(Where, x̄ is average weight of 10 walleye.)

We know that if X ~ N(μ, σ²) then, x̄ ~ N(μ, σ²/n).

And if x̄ ~ N(μ, σ²/n) then,

Z T- o/vn N(0,1)

Sample size (n) = 10

T - 12 - ::PT > 12) = P o/vn oln

::PT > 12) = P(Z > 12 – 11 2.5/10

.. Pū> 12) = P(Z > 1.2649

Using "pnorm" function of R we get, P(Z > 1.2649) = 0.1030

.:P (T> 12) = 0.1030

Hence, the required probability is 0.1030.

7) We have to find P(Σx_{​​​​​i} > 100).

100 P (Σε: > 100) = P(HΣ t,Σ n

100) P (Σε: > 100) 100) = P Ρ|> 8

Ρ (Σ; > 100) = P(x > 12.5)

We know that if X ~ N(μ, σ²) then, x̄ ~ N(μ, σ²/n).

And if x̄ ~ N(μ, σ²/n) then,

Z T- o/vn N(0,1)

Sample size (n) = 8

T- 12.5 - ::PT > 12.5) = P o/vn o/vn

::PT > 12.5) = P(Z > 12.5 - 11 2.5/18

:: P2 > 12.5) = P(Z > 1.6971

Using "pnorm" function of R we get, P(Z > 1.6971) = 0.0448

.::PT > 12.5) = 0.0448

P Η (Σ; > 100 0.0448

Hence, the required probability is 0.0448.

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