So, is 10 times as much as 3,000

A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week. The data show t

Cloud [144]

Answer:

Chi-square = 4.600

Degrees of freedom= 6

there is no evidence to reject hypothesis.

Step-by-step explanation:

H₀ : uniformly distributed

H₁: not uniformly distributed

Suppose some value of observed or expected frequency

Observed frequency : 39 40 30 40 41 49 41 Total: 280

Expected frequency : 40 40 40 40 40 40 40 Total: 280

Calculated the chi-square value

Chi-square = sigma {(O-E)^2/E}

= 0.025+0.000+2.500+0.000+0.025+2.025+0.025

4.600

Degrees of freedom=d f = n-1 = 7-1 = 6

Chi-square for 6 d.f at alpha = 0.05 is 12.59

Chi-square of calculated value < its critical value at alpha = 0.05, we see that there is no evidence to reject hypothesis.

The frequency of Sat is wrongly taken as 41 instead of 51

8 0

3 years ago

Among all monthly bills from a certain credit card company, the mean amount billed was $465 and the standard deviation was $300.

Fynjy0 [20]

Answer:

0.02% probability that the average amount billed on the sample bills is greater than $500.

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:

$\mu = 465, \sigma = 300, n = 900, s = \frac{300}{\sqrt{900}} = 10$ .