According to a study, the probability that a randomly selected teenagar shopped at a mall at least once during a week was 0.61. Let X be the number of students in a randomly selected group of 50 that will shop at a mall during the next week. (a) Compute the expected value and standard deviation of X. expected value standard deviation (b) Fill in the missing quantity. (Round your answer to the nearest whole number.)There is an approximately 2.5% chance that or more teenagers in the group will shop at the mall during the next week.

Step-by-step explanation:

Given that:

p = 0.61

If X is the the number of students in a randomly selected group  of a sample size n = 50

The expected value and the standard deviation can be computed as follows:

The expected value E(X) = np

The expected value E(X) =  50 × 0.61

The expected value E(X) = 30.5

The required standard deviation =

The required standard deviation =

The required standard deviation =

The required standard deviation =

The required standard deviation = 3.4489

The required standard deviation =  3.45

(b) Fill in the missing quantity. (Round your answer to the nearest whole number.)

There is an approximately 2.5% chance that _____ or more teenagers in the group will shop at the mall during the next week.

From the given information:

Now, we can deduce that:

the mean = 30.5

standard deviation = 3.45

Using the empirical rule:

At 95% confidence interval;

[μ - 2σ,  μ + 2σ] = [ 30.5 - 2(3.45) , 30.5 + 2(3.45)]

[μ - 2σ,  μ + 2σ] =  [ 30.5 - 6.9 , 30.5 + 6.9]

[μ - 2σ,  μ + 2σ] = [ 23.6, 37.4]

The 2.5% of the observations are less than 95% confidence interval and 2.5% observations are greater than 95% confidence interval.

The required number of teenagers is = the upper limit of the 95% confidence interval = 37

There is an approximately 2.5% chance that __37___ or more teenagers in the group will shop at the mall during the next week.

dhhghghghhrhbb it 5

d

step-by-step explanation:

probably .5

i can on others if necessary.