Answer:
a. 68% of the workers will earn between $47300 and $69700.
b. 2.5% of workers will earn above $89000
c. Approximately 0
Step-by-step explanation:
The standard normal distribution curve in the attached graph is used to solve this question.
a. The value $47300 is a standard deviation below the mean i.e. 58500-11200=47300. While $69700 is a standard deviation above the mean. I.e. 58500+12000=69700.
Between the first deviation below and above the mean, you have 34%+34%=68% of the salary earners within this range. So we have 68%of staffs earning within this range
b. The second standard deviation above the mean is $80900. i.e. 58500+11200+11200=$80900
We have 50%+13.5%+2.5%= 97.5% earning below $80900. Therefore, 100-97.5= 2.5% of the workers earn above this amount.
c. From the Standard Deviation Rule, the probability is only about (1 -0 .997) / 2 = 0.0015 that a normal value would be more than 3 standard deviations away from its mean in one direction or the other. The probability is only 0.0002 that a normal variable would be more than 3.5 standard deviations above its mean. Any more standard deviations than that, and we generally say the probability is approximately zero.
Answer:
The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).
Step-by-step explanation:
Intern No. of Breast
Number Exams Performed X²
1 30 900
2 40 1600
3 8 64
4 20 400
5 26 676
6 35 1225
7 35 1225
8 20 400
9 25 625
<u>10 20 400 </u>
<u> </u><u> ∑ 259 ∑ 7515</u>
Mean= X`= ∑x/n= 259/10= 25.9
Variance = s²= 1/n-1[∑X²- (∑x)²/n]
= 1/0[7515- (259)²/10]= 1/9[7515- 6708.1]
= 806.9/9=89.655= 89.66
Standard Deviation= √89.655= 9.4687
Hence
The value of t with significance level alpha= 0.05 and 9 degrees of freedom is t(0.025,9)= 2.262
The 95 % Confidence interval is given by
x`±t(∝,n-1) s/√n
So Putting the values
25.9± 2.262( 9.4687/√10)
= 25.9 ±2.262 (2.9943)
= 25.9 ± 6.7730
= 25.9 +6.7730=32.6730
25.9 -6.7730= 19.1269
= 19.1269, 32.6730
The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).
Answer:
Probability that 32 or more from this sample used Internet Explorer as their browser is 0.9015.
Step-by-step explanation:
We are given that according to Net Market Share, Microsoft's Internet Explorer browser has 53.4% of the global market.
A random sample of 70 users was selected.
Let 
= <u><em>sample proportion of users who used Internet Explorer as their browser.</em></u>
The z score probability distribution for sample proportion is given by;
Z = 
~ N(0,1)
where, p = population proportion of users who use internet explorer = 53.4%
= sample proportion = 
= 0.457
n = sample of users = 70
Now, probability that 32 or more from this sample used Internet Explorer as their browser is given by = P( 
0.457)
P( 0.457) = P( 
) = P(Z -1.29)
= P(Z 
1.29) = <u>0.9015</u>
The above probability is calculated by looking at the value of x = 1.29 in the z table which has an area of 0.9015.
Answer:
b
Step-by-step explanation:
Answer:
46,189
Step-by-step explanation:
The prime numbers that are less than 20 are :
1,2,3,5,7,11,13,17,19
to get the greatest value, we multiply the four numbers with the largest values i.e
11 x 13 x 17 x 19 = 46,189
