0.84(because 20% of the observations are about 0.84)
You've found the 80th percentile! The 80th percentile of the standard normal distribution is 0.84. That's because 80% of the observations fall below 0.84. (Note: The 80th percentile of every normal distribution is 0.84 times the standard deviation above the mean.)

7 0

3 years ago

Which is the graph of the function f(x) = -√x

iris [78.8K]

This is the graph, I hope it helps

3 0

2 years ago

A study is done to determine if students in the California state university system (sample 1) take longer to graduate, on averag

Debora [2.8K]

Answer:

We conclude that the students in the California state university system take longer to graduate as compared to students enrolled in private universities.

Step-by-step explanation:

We are given that from years of research, it is known that the population standard deviations are 1.5 years and 0.9 years, respectively.

One hundred students from each the California state university system and private universities are surveyed. The California state university system students took on average 4.8 years while the private university students took on average 4.4 years.

Let $\mu_1$ = population average time taken by students in the California state university system to graduate

$\mu_2$ = population average time taken by students in the private universities to graduate

SO, Null Hypothesis, $H_0$ : $\mu_1-\mu_2\leq0$ or $\mu_1 \leq \mu_2$ {means that the students in the California state university system take less than or equal time to graduate as compared to students enrolled in private universities}

Alternate Hypothesis, $H_A$ : $\mu_1-\mu_2 >0$ or $\mu_1>\mu_2$ {means that the students in the California state university system take longer to graduate as compared to students enrolled in private universities}

The test statistics that will be used here is Two-sample z test statistics as we know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1} + \frac{\sigma_2^{2} }{n_2}} }$ ~ N(0,1)

where, $\bar X_1$ = sample average time taken by the California state university system students to graduate = 4.8 years

$\bar X_2$ = sample average time taken by the private university students to graduate = 4.4 years

$\sigma_1$ = population standard deviation for California state university system students = 1.5 years

$\sigma_2$ = population standard deviation for private university students = 0.9 years

$n_1$ = sample of California state university system students taken = 100

$n_2$ = sample of private university students taken = 100

So, test statistics = $\frac{(4.8-4.4)-(0)}{\sqrt{\frac{1.5^{2} }{100} + \frac{0.9^{2} }{100}} }$

= 2.287

Now at 0.10 significance level, the z table gives critical value of 1.2816 for right-tailed test. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the students in the California state university system take longer to graduate as compared to students enrolled in private universities.

5 0

4 years ago

DELTA MATH!!!!!!!!!!!!!!!!!!!!!!!!1

Nostrana [21]

Answer:

Step-by-step explanation:

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5 0

2 years ago