A candlestick contains 5 pounds of silver. Suppose silver is worth $15 per ounce.

Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean len

Marina86 [1]

Answer:

Step-by-step explanation:

Hello!

You need to construct a 95% CI for the population mean of the length of engineering conferences.

The variable has a normal distribution.

The information given is:

n= 84

x[bar]= 3.94

δ= 1.28

The formula for the Confidence interval is:

x[bar]± $Z_{1-\alpha/2}$ *(δ/n)

Lower bound(Lb): 3.698

Upper bound(Ub): 4.182

Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242

I hope it helps!

6 0

2 years ago

Find the total cost to the nearest cent $43 dinner 18% gratuity

Norma-Jean [14]

43*1.18=$50.74
You pay 100% plus 18% tip

5 0

2 years ago

The Big River Casino is advertising a new digital lottery-style game called Instant Lotto. The player can win the following mone

zysi [14]

Answer:

(a) The expected value of the prize for one play of Instant Lotto is $3.50.

(b) The probability that the visitor wins some prize at least twice in the 20 free plays is 0.2641.

(c) The probability that a randomly selected day has at least 1000 people play Instant Lotto is 0.2579.

Step-by-step explanation:

(a)

The probability distribution of the monetary prizes that can be won at the game called Instant Lotto is:

X P (X = x)

$10 0.05

$15 0.04

$30 0.03

$50 0.01

$1000 0.001

$0 0.869

___________

Total = 1.000

Compute the expected value of the prize for one play of Instant Lotto as follows:

$E(X)=\sum x\cdot P (X=x)$

$=(10\times 0.05)+(15\times 0.04)+(30\times 0.03) \\+ (50\times 0.01)+(1000\times 0.001)+(0\times 0.869)\\=0.5+0.6+0.9+0.5+1+0\\=3.5$

Thus, the expected value of the prize for one play of Instant Lotto is $3.50.

(b)

Let X = number of times a visitor wins some prize.

A visitor to the casino is given n = 20 free plays of Instant Lotto.

The probability that a visitor wins at any of the 20 free plays is, p = 1/20 = 0.05.

The event of a visitor winning at a random free play is independent of the others.

The random variable X follows Binomial distribution with parameters n = 20 and p = 0.05.

Compute the probability that the visitor wins some prize at least twice in the 20 free plays as follows:

P (X ≥ 2) = 1 - P (X < 2)

= 1 - P (X = 0) - P (X = 1)

$=1-[{20\choose 0}0.05^{0}(1-0.05)^{20-0}]-[{20\choose 1}0.05^{1}(1-0.05)^{20-1}]\\=1-0.3585-0.3774\\=0.2641$

Thus, the probability that the visitor wins some prize at least twice in the 20 free plays is 0.2641.

(c)

Let X = number of people who play Instant Lotto each day.

The random variable X is normally distributed with a mean, μ = 800 people and a standard deviation, μ = 310 people.

Compute the probability that a randomly selected day has at least 1000 people play Instant Lotto as follows:

Apply continuity correction:

P (X ≥ 1000) = P (X > 1000 + 0.50)

= P (X > 1000.50)

$=P(\frac{X-\mu}{\sigma}>\frac{1000.50-800}{310})$

$=P(Z>0.65)\\=1-P(Z$

Thus, the probability that a randomly selected day has at least 1000 people play Instant Lotto is 0.2579.

5 0

2 years ago

The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in n

devlian [24]

Answer:

The length of the longest 15% of Atlantic cod in this area is 53.79cm, roundeed to 2 decimal places.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have that:

A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm, so $\mu = 49.9, \sigma = 3.74$ .

What is the length in cm of the longest 15% of Atlantic cod in this area?

We have to find the value of X for the value of Z that has a pvalue of 0.85.

Looking at the zscore table, we have that Z = 1.04 has a pvalue of 0.8508. So, we have to find the value of X when $Z = 1.04, \mu = 49.9, \sigma = 3.74$ .