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torisob [31]
2 years ago
13

The weight of a soccer ball is normally distributed with a mean of 21 oz and a standard deviation of 3 oz. Suppose 1000 differen

t soccer balls are in a warehouse. About how many soccer balls weigh more than 24 oz
A. 40

B. 80

C. 160

D. 200
Mathematics
2 answers:
horsena [70]2 years ago
8 0

Answer:

The correct option is C.

Step-by-step explanation:

Given information: The weight of a soccer ball is normally distributed with a mean of 21 oz and a standard deviation of 3 oz. Total number of soccer balls is 1000.

We have to find the number of soccer balls having weight more than 24 oz.

Probability of the ball having weight more than 24 oz is

P(x>24)=P(\frac{x-\mu}{\sigma}>\frac{24-21}{3})

P(x>24)=P(z>1)

P(x>24)=1-P(z\leq 1)

P(x>24)=1-P(z\leq 1)                  (Using standard normal table)

P(x>24)=1-0.8413

P(x>24)=0.1587

The number of soccer balls having weight more than 24 oz is

1000\times P(x>24)=1000\times 0.1587\Rightarrow 158.7\approx 160

The number of soccer balls having weight more than 24 oz is about 160.

Therefore the correct option is C.

Novosadov [1.4K]2 years ago
3 0

A weight of 24 oz is 1 standard deviation above the mean 21 oz. The empirical rule says approximately 68% of soccer balls have weights within 1 standard deviation (i.e. between 18 and 24 oz), so the remaining 32% of balls have weights below 18 oz or above 24 oz. The balls' weights are normally distributed, so the percentage of balls with weight less than 18 oz is the same as the percentages of balls with weight greater than 24 oz, which means about 16% of balls have weight greater than 24 oz. 16% of 1000 is 160.

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4. Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln
pochemuha

Answer:

No these these result do not differ at 95% confidence level  

Step-by-step explanation:

From the question we are told that

  The first concentrations is  c _1= 30.0 \ g/m^3

      The second concentrations is  c _2 = 52.9 \ g/m^3

  The first sample size is  n_1 =  32

    The second sample size is  n_2 =  32

   The  first standard deviation is \sigma_1 =  30.0

     The  first standard deviation is \sigma_1 =  29.0

The mean for  Turnpike is  \= x _1 = \frac{c_1}{n}  =  \frac{31.4}{32} = 0.98125

The mean for   Tunnel is  \= x _2 = \frac{c_2}{n}  =  \frac{52.9}{32} = 1.6531

The  null hypothesis is  H_o  :  \mu _1 - \mu_2  =  0

The  alternative hypothesis is  H_a  :  \mu _1 - \mu_2  \ne  0

Generally the test statistics is mathematically represented as

              t =  \frac{\= x_1 - \= x_2}{ \sqrt{\frac{\sigma_1^2}{n_1}  +\frac{\sigma_2^2}{n_2} }}

         t =  \frac{0.98125 - 1.6531}{ \sqrt{\frac{30^2}{32}  +\frac{29^2}{32} }}

        t = - 0.0899

Generally the degree of freedom is mathematically represented as

     df =  32+ 32 - 2

     df =  62

The  significance \alpha  is  evaluated as

      \alpha  =  (C - 100 )\%

=>   \alpha  =  (95 - 100 )\%

=>   \alpha  =0.05

The  critical value  is evaluated as

      t_c  =  2  *  t_{0.05 ,  62}

From the student t- distribution table  

        t_{0.05, 62} =  1.67

So

     t_c  =  2 * 1.67

=>  t_c  = 3.34

given that

       t_c  >  t we fail to reject the null hypothesis so  this mean that the result do  not differ

       

6 0
1 year ago
Which score has a higher relative position, a score of 271.2 on a test for which = 240 and , or a score of 63.6 on a test for wh
Fittoniya [83]

Answer:

The score of 271.2 on a test for which xbar = 240 and s = 24 has a higher relative position than a score of 63.6 on a test for which xbar = 60 and s = 6.

Step-by-step explanation:

Standardized score, z = (x - xbar)/s

xbar = mean, s = standard deviation.

For the first test, x = 271.2, xbar = 240, s = 24

z = (271.2 - 240)/24 = 1.3

For the second test, x = 63.6, xbar = 60, s = 6

z = (63.6 - 60)/6 = 0.6

The standardized score for the first test is more than double of the second test, hence, the score from the first test has the higher relative position.

Hope this Helps!!!

3 0
1 year ago
A fish tank is three-quarters (3/4) full. two-thirds (2/3) of the water leaks out, leaving 10 gallons of water in the tank. what
Naily [24]

To solve this problem, all we have to do is to thoroughly analyze the situation step by step.

First let us work on the remaining 10 gallons of water in the tank. We are given that 2 / 3 of the water initially in the tank leaked out. Therefore this means that we are left with 1 / 3 the original amount of water which is 10 gallons, hence:

(1 / 3) Vi = 10 gallons

where Vi is the initial volume of water inside the tank, calculating for Vi:

Vi = 3 (10 gallons)

Vi = 30 gallons

 

We are also given that this initial volume of water is just 3 / 4 of the total capacity of the tank, therefore:

(3 / 4) Vc = Vi

where Vc is the volume capacity of the tank, calculating for Vc:

Vc = 4 Vi / 3

Vc = 4 (30 gallons) / 3

Vc = 40 gallons

 

Answer:

<span>Capacity of the fish tank is 40 gallons</span>

3 0
2 years ago
the average number of spectators at a football competition for the first five days was 3144.the attendance on the sixth day was
Ksivusya [100]

Answer:

3285

Step-by-step explanation:

5x3144=15720

3990

15720+3990=19710

19710/6=3285

8 0
2 years ago
The table below shows the total cost of buying printer ink cartridges for both members and nonmembers of a shopping club. Price
ki77a [65]

Answer:

B

Step-by-step explanation:

3 0
2 years ago
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