A survey of several 9 to 11 year olds recorded the following amounts spent on a trip to the mall: $10.31,$17.22,$26.62,$22.84 Construct the 98% confidence interval for the average amount spent by 9 to 11 year olds on a trip to the mall. Assume the population is approximately normal. Step 1 of 4 : Calculate the sample mean for the given sample data. Round your answer to two decimal places.

Step-by-step explanation:

From the information given,

Mean, μ = (10.31 + 17.22 + 26.62 + 22.84)/4 = 19.2475

Standard deviation, σ = √summation(x - mean)/n

Summation(x - mean) = (10.31 - 19.2475)^2+ (17.22 - 19.2475)^2 + (26.62 - 19.2475)^2 + (22.84 - 19.2475)^2 = 151.249475

σ = √(151.249475/4)

σ = 6.15

number of sample, n = 4

The z score for 98% confidence interval is 2.33

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

19.2475 ± 2.33 × 6.15/√4

= 19.2475 ± 2.33 × 3.075

= 19.2475 ± 7.16

The lower end of the confidence interval is 19.2475 - 7.16 = 12.09

The upper end of the confidence interval is 19.2475 + 7.16 = 26.41

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

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