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eduard
2 years ago
13

The distribution of number of hours worked by volunteers last year at a large hospital is approximately normal with mean 80 and

standard deviation 7. Volunteers in the top 20 percent of hours worked will
receive a certificate of merit. If a volunteer from last year is selected at random, which of the following is closest to the probability that the volunteer selected will receive a certificate of merit given that the
number of hours the volunteer worked is less than 90?
A 0.077
B 0.123
C 0 0.618
D 0.618
E 0.923
Mathematics
1 answer:
wariber [46]2 years ago
7 0

Answer:

E 0.923

Step-by-step explanation:

The Z score is a measurement used in statistics to determine the relationship between a value and a mean of a group of values measured as standard deviation from the mean. The z score (z) is given as:

z=\frac{x-\mu}{\sigma}, Where μ is the mean, σ is the standard deviation and x is the value.

Given that:

σ = 7, μ = 80 and x = 90  

z=\frac{x-\mu}{\sigma}=\frac{90-80}{7} =1.43

From the normal distribution table, P(X < 90) = P(z < 1.43) = 0.923

You might be interested in
Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s
stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.  

Then we have n_{1} = 25, \bar{x}_{1} = 2.04, \sigma_{1} = 0.010 and n_{2} = 20, \bar{x}_{2} = 2.07, \sigma_{2} = 0.015. The pooled estimate is given by  

\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative).  

The test statistic is T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}} and the observed value is t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257. T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value t_{0} falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by 2P(T0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}, i.e.,

-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}

where t_{0.025} is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

-0.03\pm(2.0167)(0.012459)(0.3), i.e.,

(-0.0225, -0.0375)

8 0
2 years ago
Find the function rule.<br><br> X: -2, -1, 0, 1, 2<br> Y: 9,4, -1, -6, -11
8090 [49]
X: +1
y: -5
..,.........
5 0
2 years ago
If f(x) = 5x + 40, what is f(x) when x = –5?<br><br> a.–9<br> b.–8<br> c.7<br> d.15
defon

Just plug the number

_5(5)+40=-25+40

=15

3 0
2 years ago
Read 2 more answers
Jared is an intern at a real estate broker’s office. He was asked to record data on the difference of the number of sales made e
polet [3.4K]

Answer:

1.)

interval over which the difference of the numbers of sales is increasing: (<u>4.9, 10)</u>

interval over which the difference of the numbers of sales is decreasing: <u>(0, 4.9)</u>

interval over which the difference of the numbers of sales is positive: <u>(-9, -3)</u>

interval over which the difference of the numbers of sales is negative: <u>(-3, 10)</u>

<u />

2.)

The team’s total number of sales is equal to the average number of sales at

<u>-9, -3, and 10 </u>months from when Jared began gathering data.

The team’s minimum number of sales this year occurred at approximately

<u>-28</u> months from when Jared began gathering data.

3.)

(x + 3)

(x + 9)

(x - 10)

4.)

f(x) = 0.05x^3 + 0.1x^2 - 4.65x - 13.5

5.)

a = 0.05

b = -9

c = -3

d = 10

Step-by-step explanation:

3 0
2 years ago
Mr Suray had some money. He spent $154 on a jacket and 3/8 of the remainder on a shaver. He had 1/3 of his money left.
nadya68 [22]
(a) What fraction of the money did I spend on the jacket?
 For this case the first thing you should know is that all the money is represented in the following way:
 100% = 1
 Therefore, we write the following equation:
 x + 3/8 + 1/3 = 1
 From here, we clear x:
 x = 1- 3/8 - 1/3
 x = 7/24
 Answer: 
 he did spend on the jacket 7/24 of the money

 (b) How much money did he have at first?
 We can make the following rule of three:
 7/24 ----> $ 154
 1 --------> x $
 We clear the value of x:
 x = (1 / (7/24)) * (154)
 x = $ 528
 Answer:
 he had at first 528 $
4 0
2 years ago
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