Assume that the weight of hamsters is Normally distributed with μ = 1.2 lbs and σ = 0.25 lbs. Suppose I randomly select 4 hamsters.

(a) What is the shape of the sampling distribution of the sample mean? (unknown, exactly normal, approximately normal)

(b) What is P(X < 1.14)? (Use 3 decimal places)

Given Information:  

Population mean = μ = 1.2

Population standard deviation = σ = 0.25  

Sample size n = 4

Required Information:  

P(X < 1.14) = ?

The probability of P(X < 1.14) is 0.315

Explanation:  

(a) What is the shape of the sampling distribution of the sample mean?

Since it is given that population is normally distributed then sample distribution will also be normally distributed even with a smaller sample size .

When population is not normally distributed and is very skewed then a larger sample size is required to ensure normal sample distribution according to central limit theorem.

(b) What is P(X < 1.14)?

First let us find out sample mean and standard deviation

sample mean will be same as population mean μ = 1.2

sample standard deviation =  σ/√n

where n = 4 is given

sample standard deviation =  σ/√4 =  0.125

Now we can calculate the probability using z-score

P(X < 1.14) = P(Z < (1.14 - 1.2)/0.125) = P(Z < -0.48)

From the z-table

P(Z < -0.48) = 0.3156

Therefore, the probability of P(X < 1.14) is 0.315 or 31.5%

first one because it can't have a fraction in it

step-by-step explanation:

c = 0.429t

step-by-step explanation:

the constant rate at which carlos harvests cassava is 35 minutes per 15 cassavas 15/35 = 0.429 cassava per minute

therefore, an equation that models the number of cassava (c) that harvest carlos in a time (t) is the following:

where t is the time in minutes.

this means that when 35 minutes have elapsed 15 cassava have been harvested, at 70 minutes 30 cassava have been harvested and so on.