Ask question

Ask question

All categories

2 years ago

6

"the placement test for a college has scores that are normally distributed with a mean of 600 and a standard deviation of 60.if

the college accepts only the top 1% of examinees, what is the cutoff score on the test for admission?"

1 answer:

Tanzania [10]2 years ago

4 0

The cutoff will be evaluated as follows:
let the cutoff be x

but
P(x>0.01)=1-P(x<0.01)=1-0.01=0.99
the z value corresponding to 0.99% is z=2.33
thus
2.33=(x-600)/60
solving for x we get:
139.8=x-600
x=739.8

You might be interested in

Question 14(1 point Find x. Round answer to the nearest tenth. (one decimal place - if you round incorrectly, the answer will be

Volgvan

Answer:

LOL BRO ENGLISH DO U SPEAK IT

8 0

2 years ago

I need a little help. (The question is on the screenshot by the way)

vagabundo [1.1K]

Answer:

45

Step-by-step explanation:

the answer iud D

7 0

2 years ago

Read 2 more answers

What is 5 (p-10) expanded

V125BC [204]

Answer:

5p - 50

You distribute

5 x p
5 x -10

3 0

2 years ago

Read 2 more answers

Plantea una ecuacion

Anna35 [415]

Answer:78phy

gigh[ood9..8d55.5.5.5.5.5.5.5.6.s7s47.4

Step-by-step explanation:

8 0

3 years ago

5) In a certain supermarket, a sample of 60 customers who used a self-service checkout lane averaged 5.2 minutes of checkout tim

Ne4ueva [31]

Answer:

$\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643$

$S_p=2.940$

$t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751$

$df=60+72-2=130$

$p_v =P(t_{130}$

Assuming a significance level of $\alpha=0.05$ we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

Step-by-step explanation:

Data given

Our notation on this case :

$n_1 =60$ represent the sample size for people who used a self service

$n_2 =72$ represent the sample size for people who used a cashier

$\bar X_1 =5.2$ represent the sample mean for people who used a self service

$\bar X_2 =6.1$ represent the sample mean people who used a cashier

$s_1=3.1$ represent the sample standard deviation for people who used a self service

$s_2=2.8$ represent the sample standard deviation for people who used a cashier

Assumptions

When we have two independent samples from two normal distributions with equal variances we are assuming that

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

The statistic is given by:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

And t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

System of hypothesis

Null hypothesis: $\mu_1 \geq \mu_2$

Alternative hypothesis: $\mu_1 < \mu_2$

This system is equivalent to:

Null hypothesis: $\mu_1 - \mu_2 \geq 0$

Alternative hypothesis: $\mu_1 -\mu_2 < 0$

We can find the pooled variance:

$\S^2_p =\frac{(60-1)(3.1)^2 +(72 -1)(2.8)^2}{60 +72 -2}=8.643$

And the deviation would be just the square root of the variance:

$S_p=2.940$

The statistic is given by:

$t=\frac{(5.2 -6.1)-(0)}{2.940\sqrt{\frac{1}{60}+\frac{1}{72}}}=-1.751$

The degrees of freedom are given by:

$df=60+72-2=130$

And now we can calculate the p value with:

$p_v =P(t_{130}$

Assuming a significance level of $\alpha=0.05$ we have that the p value is lower than this significance level so then we can conclude that the mean for checkout time is significantly less for people who use the self-service lane

5 0

3 years ago