The following data represent the​ high-temperature distribution for a summer month in a city for some of the last 130 years. Treat the data as a population. Complete parts​ (a) through​ (c).

Temperature Lower Limit Upper Limit Days
50-59 50 59 2
60-69 60 69 313
70-79 70 79 1419
80-89 80 89 1503
90-99 90 99 319
100-109 100 109 9

Required:
Approximate the mean and standard deviation for temperature.

Answer:

The mean is  [tex]\= x  = 79.7 [/tex]

The standard deviation  [tex]\sigma =7.81[/tex]

Step-by-step explanation:

From the question we are told that

  The data given is  

Temperature                Lower Limit           Upper Limit               Days

50-59                               50                             59                          2

60-69                                 60                            69                        313

70-79                                  70                           79                          1419

80-89                                   80                          89                         1503

90-99                                   90                          99                          319

100-109                                100                         109                         9        

Generally the average of each limit is evaluated as

Temperature    Lower Limit   Upper Limit         Average[tex](x_i)[/tex]                    Days[tex](f_i)[/tex]

50-59                 50                   59              ( 50 + 59 )/2 =54.5                 2

60-69                 60                   69              (60+69)/2 = 64.5                    313

70-79                 70                    79               (70+79)/2 =74.5                    1419

80-89                 80                   89                (80+89)/2 =84.5                  1503

90-99                 90                   99                 (90+99)/2 = 94.5                319

100-109             100                  109                 (100+109)/2 = 104.5            9  

Generally the mean is  evaluated as

     [tex]\mu  =  \frac{\sum x_i f_i}{\sum f_i}[/tex]

=>  [tex]\mu  =  \frac{ (54.5 * 2) + (64.5 * 313) +\cdots +(104.5 *  9)}{2 + 313 +\cdots + 9}[/tex]

=> [tex]\= x  = 79.7 [/tex]

Generally the standard deviation  is  evaluated as

    [tex]\sigma = \sqrt{ \frac{\sum[ f_i * ( x_i^2 - \= x^2)] }{\sum f_i}}[/tex]

=>   [tex]\sigma =\sqrt{ \frac{[2 * [(54.5)^2) - 79.7^2 ]]+[313 *[ (64.5)^2)-79.7^2] +\cdots +[9 *[ (104.5)^2)- 79.7^2]  }{2 + 313 +\cdots + 9}}[/tex]

=>  [tex]\sigma =7.81[/tex]

The mean and Standard deviation for the given data would be as follows:

μ [tex]= 79.7[/tex]

б [tex]= 7.81[/tex]

Given that,

Temperature       Lower Limit           Upper Limit             Days([tex]f_{i}[/tex])

50-59                       50                           59                          2

60-69                       60                           69                         313

70-79                         70                           79                          1419

80-89                        80                          89                          1503

90-99                        90                          99                           319

100-109                    100                         109                            9  

Average([tex]x_{i}[/tex])    

[tex]( 50 + 59 )/2 =54.5[/tex]

[tex](60+69)/2 = 64.5[/tex]

[tex](70+79)/2 =74.5[/tex]

[tex](80+89)/2 =84.5[/tex]

[tex](90+99)/2 = 94.5[/tex]

[tex](100+109)/2 = 104.5[/tex]

As we know,

Mean (μ) [tex]=[/tex] ∑[tex](x_{i} f_{i}) /[/tex]∑[tex]f_{i}[/tex]

[tex]= [(54.5[/tex] × [tex]2) + (64.5[/tex] ×[tex]313)+....(104.5[/tex]× [tex]9)][/tex][tex]/(2 + 313 + 1419 + ....9)[/tex]

∵ μ [tex]= 79.7[/tex]

As we know,

Standard Deviation:-

[tex]\sigma={\sqrt {\frac {\sum(x_{i}-{\mu})^{2}}{f_{i} }}}[/tex]

by putting the values, we get

∵ б [tex]= 7.81[/tex]

Thus, μ [tex]= 79.7[/tex] and б [tex]= 7.81[/tex] are the correct answers.

Learn more about "Temperature" here:

brainly.com/question/7510619

is 3.4567 irrational or rtional

is 1.05 a terminating decimal

A fashion designer wants to know how many new dresses women buy each year. Assume a previous study found the standard deviation to be 1.3. He thinks the mean is 5.5 dresses per year. What is the minimum sample size required to ensure that the estimate has an error of at most 0.1 at the 95% level of confidence? Round your answer up to the next integer.

Which inequality is true?

Simplify the algebraic expression by combining like (or similar) terms.
5+9x+2

Help me please I really need help

what is the answer the question 3? pls help!!

i mark as brainliest ​

solve the following equation for a: -g + 3/4a = y

Find the area and perimeter of a rectangle whose sides are √2 + 3 √5 + 2

whats 62.7×2.45......?

Evaluate 8 ÷ -2 · 4²+9. Thanks! Will give brainiest if possible

What is the slope of the equation of this line? (Put into y=mx+b form first). 6x+y=-1

Trevor paints 1/6 of the fence surrounding his farm each day how many days will it take him to paint 3/4 of the fence?

What operation is happening in an additive pattern??

Multiplication
Subtraction
Addition
Division

Identify the holes, vertical asymptotes, and horizontal asymptotes of the following.


[tex]f (x) = \frac{2}{x-3}[/tex]

Alice has x dollars in her bank account. She deposits $75 in her account. After the deposit, the account has $550. Write an equation that could be used to figure out how much money was in her account before the deposit.

I need help I don't understand ​

Find the slope of the line y=-3x+3

How many sandwiches must you sell to make $100 profit

Your economics instructor assigns you and a student you do not know a joint project. Assume that you and your partner are both interested in maximizing your grade, but you are both very busy and get more happiness if you can get a good grade with less work. Since you do not know the student, making a credible commitment to how much you are going to work is impossible. Note that on the payoff matrix below, 10/10 is happiest and 1/10 is least happy. This level of happiness is a combination of how much you work and the grade you achieve. Assume that you will have to work on only one project with this partner.
You Work Hard Work Less Hard Grade-A, but you had to work 6 hours. Happiness-6/10 [Grade A, and you only worked 2 hours. Happiness-10/10 Your Partner Work Har Grade-A, but your partner had to work 6 hours. Happiness-6/10 Grade-A, but your partner had to work 10 hours. Happiness 4/10 Work Less Hard Grade-A, but you had to work 10 hours. Happiness-3/10 Grade B, but you only worked 4 hours. Happiness 5/10 Grade A, and your partner only worked 2 hours. Happiness-8/10 Grade- B, but your partner only worked 4 hours. Happiness-5/10 In this game, both you and your partner will collectively work a total of hours on the project.

A coin is about to be tossed 50 times. Determine the probability the 8th heads occurs on the 12th toss

What is the inverse of the function? *
f(x)=3x+1

Aniya is helping her father paint a fence. She can paint 2 sections of the fence in 15 minutes. How many minutes will it take Aniya to paint 12 sections of the fence.

Given parallelogram ABCD, the measure of angle C

What is a reasonable estimate of -3 4/5 + (-5.25) and the actual value?​

how to construct cuboid net? what should be the measurement (around 1-5)?
​

How many milliters are in 90 ounces

The correct Standard Form of the equation y=−2/7x−11/7

iiii need heeelllppp with this one