Answer:
The mean of W is 55 ounces.
The standard deviation of W is 4.33 ounces.
Step-by-step explanation:
Let X: weight of a red delicious apple, and B: the weight of the box and packing material.
The distribution that will represent W: the total weight of the packaged 5 randomly selected apples will be also normally distributed.
Applying the property of the mean: [tex]E(aX)=aE(X)[/tex], the mean of W will be:
[tex]\mu_W=E(W)=E(5X+B)=5E(X)+E(B)=5*9+10=45+10=55[/tex]
Applying the property of the variance: [tex]V(aX+b)=a^2V(X)[/tex], the variance of W will be:
[tex]\sigma^2_W=V(W)=V(5X+B)=5^2V(X)+V(B)=25*0.75+0=18.75[/tex]
The mean standard deviation of W will be the squared root of V(W):
[tex]\sigma_W=\sqrt{V(W)}=\sqrt{18.75}=4.33[/tex]
The mean of W is 55 ounces.
The standard deviation of W is 4.33 ounces.
The distribution in this question is a normal distribution. The calculated mean is 55 and the probability is 0.2219.
The distribution of the total weight of the randomly selected apples is a normal distribution.
Mean = 5 x 9 =45
standard deviation = 5 x 0.75 = 3.75
The distribution would then be T ~ N(μ = 45, σ = 3.75)
2. probability = can be solved by:
probability (T<42)
p(z < 42-45)/3.75)
prob(z <-0.8) = 0.2119 is the required probability.
w = 10+T = 10+5X
mean (w) = 10 + 45 = 55
The calculated mean is therefore 55.
sd(w) = SD(t) = 3.75
Read more on probability here: https://brainly.com/question/24756209
