The table shows the number of photocopies made during one day at each of the 25 schools in a school district.

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate

Leviafan [203]

Answer:

Probability of having student's score between 505 and 515 is 0.36

Given that z-scores are rounded to two decimals using Standard Normal Distribution Table

Step-by-step explanation:

As we know from normal distribution: z(x) = (x - Mu)/SD

where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution

Therefore using given data: Mu (Mean) = 510, SD = 10.4 we have z(x) by using z(x) = (x - Mu)/SD as under:

In our case, we have x = 505 & 515

Approach 1 using Standard Normal Distribution Table:

z for x=505: z(505) = (505-510)/10.4 gives us z(505) = -0.48

z for x=515: z(515) = (515-510)/10.4 gives us z(515) = 0.48

Afterwards using Normal Distribution Tables and rounding the values to two decimals we find the probabilities as under:

P(505) using z(505) = 0.32

Similarly we have:

P(515) using z(515) = 0.68

Now we may find the probability of student's score between 505 and 515 using:

P(505 < x < 515) = P(515)-P(505) = 0.68 - 0.32 = 0.36

PS: The standard normal distribution table is being attached for reference.

Approach 2 using Excel or Google Sheets:

P(x) = norm.dist(x,Mean,SD,Commutative)

P(505) = norm.dist(505,510,10.4,1)

P(515) = norm.dist(515,510,10.4,1)

Probability of student's score between 505 and 515= P(515) - P(505) = 0.36

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6 0

3 years ago

What is the missing length?

Kipish [7]

Answer:

80

Step-by-step explanation:

89^2-39^2=6400, 6400=80^2

4 0

3 years ago

What is the mathematic problem 1 3/4 x 2

Vikentia [17]

Answer:

i believe it would be 6 1/2 simplified or 6 2/4 unsimplified

Step-by-step explanation:

8 0

3 years ago

Answer it please! Nobody at home will help me...

STALIN [3.7K]

The area formula for a triangle is 1/2 (base * height)

4 0

3 years ago

Find the critical value of x2 based on the given information H1 >3.5 n=14 a=0.05

egoroff_w [7]

Answer:

The degrees of freedom are given by:

$df = n-1=14-1=13$

The significance level is $\alpha=0.05$ and then the critical value can be founded in th chi square table we need a quantile that accumulates 0.05 of the area in the right tail of the distribution and for this case is:

$\chi^2_{\alpha}= 22.362$

And if the chi square statistic is higher than the critical value we can reject the null hypothesis in favor of the alternative.

Step-by-step explanation:

We have the followign system of hypothesis:

Null hypothesis: $\sigma^2 \leq 3.5$

Alternative hypothesis: $\sigma^2 >3.5$

The degrees of freedom are given by:

$df = n-1=14-1=13$

The significance level is $\alpha=0.05$ and then the critical value can be founded in th chi square table we need a quantile that accumulates 0.05 of the area in the right tail of the distribution and for this case is:

$\chi^2_{\alpha}= 22.362$

And if the chi square statistic is higher than the critical value we can reject the null hypothesis in favor of the alternative.

4 0

3 years ago