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ICE Princess25 [194]
3 years ago
10

In a random sample of 400 people who were questioned regarding their participation in sports, 160 said that they did participate

. find a 98% confidence interval for p. put your answers in interval notation i.e. (a,
b.and round your answers to 3 decimal places.
Mathematics
1 answer:
Scilla [17]3 years ago
8 0
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Read 2 more answers
Potatoes: Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 9 ounces and a standard deviatio
katrin2010 [14]

Answer: a)  79.10 %           b) 7%                c)19.85 %

Step-by-step explanation:

a) Z = ( × - μ ) ÷ σ           Where  μ mean of population σ standard               deviation Z is the abscissa to give the area or probability we are looking for associated to the value 10 ounces ( × )

So:  Z = ( 10 - 9 ) ÷ 1.2  ⇒ Z = 1/1.2 ⇒   Z = 0.83  it has to be below this value

From Z tables we get  P [ Z ≤ 0.82 ] = 0.7910  0r  79.10 %

b) Following the same procedure: We look for

P [ Z > ( x - μ ) ÷   σ ]    ⇒   Z  =  ( 12 -10 ) ÷ 1.2  = 2.5

From Z table we get the area under the curve from the left tail up to the point Z < 2.5 ( 2.5 not included) but we were asked for the area out of that previous so 1- 0.9930 = 0.007 is the area we are looking for

So P (b) = 0.007  or 7 %

Finally between the two points above mentioned ( 10 ≤  Z  ≤ 12 ) we use the previous values (taking in consideration the limits, according to the problem statement )

Z ≤ 10     Z ⇒( 10-9  ) ÷ 1.2   Z = 0.7967

Z ≥ 12     Z ⇒ ( 12 - 9 ) ÷ 1.2  Z = 0.9952

The interval is between these two points

0.9952 - 0.7967 = 0.1985   ⇒ or 19.85 %

The attached help in the understanding of the solution

                               

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