National League Hockey play-off games go into overtime if the score is tied at the end of a game. Overtime lasts until one of the teams scores a goal. From years of experience we know that the distribution of the lengths of the overtime periods is right skewed with a mean of 9.8 minutes and a standard deviation of 12 minutes. What is the (approximate) probability that a random sample of 30 overtime periods would have a (sample) mean length of more than 13 minutes? Give your answer to 3 decimal places, e. g, 0.123.

0.07215 = 0.072 to 3 d.p.

Step-by-step explanation:

Central limit theorem explains that the sampling distribution obtained from this distribution will be approximately a normal distribution with

Mean = population mean

μₓ = μ = 9.8 minutes

Standard deviation of the distribution of sample means = σₓ = (σ/√n)

σ = 12 minutes

n = sample size = 30

σₓ = (12/√30) = 2.191

Probability that a random sample of 30 overtime periods would have a (sample) mean length of more than 13 minutes

Required probability = P(x > 13)

Since we've established that this distribution of sample means approximates a normal distribution

We first standardize 13 minutes.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (13 - 9.8)/2.191 = 1.46

Required probability

P(x > 13) = P(z > 1.46)

We'll use data from the normal probability table for these probabilities

P(x > 13) = P(z > 1.46) = 1 - P(z ≤ 1.46)

= 1 - 0.92785 = 0.07215

Hope this Helps!!!

don't report but were is the digram

step-by-step explanation:

no because mercury has a density of 13.6 grams per centimeter squared while gold has 19.32 grams centimeter squared

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