How do you write 0.512 as a fraction

There are 42 students in an elementary statistics class. On the basis of years of experience, the instructor knows that the time

harkovskaia [24]

Answer:

A) 0.99413

B) 0.00022

Step-by-step explanation:

A) First of all let's find the total grading time from 6:50 P.M to 11:00 P.M.:

Total grading time; X = 11:00 - 6:50 = 4hours 10minutes = 250 minutes

Now since we are given an expected value of 5 minutes, the mean grading time for the whole population would be:

μ = n*μ_s ample = 42 × 5 = 210 minutes

While the standard deviation for the population would be:

σ = √nσ_sample = √(42 × 6) = 15.8745 minutes

To find the z-score, we will use the formula;

z = (x - μ)/σ

Thus;

z = (250 - 210)/15.8745

z = 2.52

From the z-distribution table attached, we have;

P(Z < 2.52) ≈ 0.99413

B) solving this is almost the same as in A above, the only difference is an additional 10 minutes to the time.

Thus, total time is now 250 + 10 = 260 minutes

Similar to the z-formula in A above, we have;

z = (260 - 210)/15.8745

z = 3.15

P(Z > 3.15) = 0.00022

5 0

3 years ago

. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an

Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$

Normal probability distribution

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:

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

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12$