Question

normally distributed) with a mean of 32 ounces and a standard deviation 1. The weights of bags of of 0.36 ounce. Bags in the upper 4.5% are too heavy and must be repackaged, what is the most a bag of baby carrots can weigh and not need to be repackaged? -5 points 2. Som e college students use credit cards to pay for school-related expenses. For this population, the amount paid is normally distributed, with a mean of $1615 and a standard deviation of $ 550. hat is the probability that a randomly selected college student, who uses a credit card to pay for a. W school-related expenses, paid less than $1400? -4 points b. You randomly select 25 college students who use credit cards to pay for school-related expenses. What is the probability that their mean amount paid is less than $1400? - 4 points Compare the probability from part 1 and 2. c.

Answer 1

1.

The following information has been provided:

μ=32, σ= 0.36

We need to compute x such that

. The corresponding z-value needed to be computed is:

Pr(X > x) 0045

$Z = \frac{X - \mu}{\sigma} = \frac{ x - 32}{ 0.36}$

Therefore, we get that

X-32 0.36 0.045

-32 = 0.36 1.6954

→ X-32.61034318

The following is obtained graphically:

Normal Distribution: Pr(X> 32.61034318) 0.045 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 30.4 30.6 30.8 31.0 31.2 31.4 31.6 31.8 32.0 32.2 32.4 32.6 32.8 33.0 33.2 33.4 33.6

2.

a.

The following information has been provided μ-1615, σ-550 We need to compute Pr( 1400). The corresponding z-value needed to be computed: 1400 -16150.3909 x-μ X μ- 1615 550 Therefore, 1400 1615 Pr(X 3 1400)-Pr (Z550 Pr(Z <-0.3909)-0.3479

b.

The following information has been provided:

μ= 1615, σ= 550/5-110

We need to compute Pr( 1400). The corresponding z-value needed to be computed: 1400 1615 110 Z =-1.9545 Therefore 1400 1615 )-Pr(z- 1400) = Pr (ZS Pr(X 0253 1.9545) .

c.

The probability in part a is higher due to higher variability. As we increase the sample size even though the mean remains the same, the variability decreases. This is causing a relatively lower chance of a value far away from the mean like 1400 to be possible. Hence the probability in part b is smaller than part a.

Let me know in comments if anything is unclear. Will reply ASAP. Please upvote!

Answer 2

For the second problem about the credit cards, why did you divide the standard deviation by 5?