Arecent study was conducted to investigate the duration of time required to complete a certain manual dexterity task. the reported mean was 10.2 seconds with a standard deviation of 16.0 seconds. suppose the reported values are the true mean and standard deviation for the population of subjects in the study. if a random sample of 144 subjects is selected from the population, what is the approximate probability that the mean of the sample will be more than 11.0 seconds?

Using the Zscore relation, the approximate probability of having a sample mean greater than 11.0 seconds is 0.2739

The standard error = σ/√n = (16 /√144)

The standard error, S. E = 16/12 = 1.33

The Zscore :

The Zscore = (11.0 - 10.2) ÷ 1.33

The Zscore = 0.601

The probability that mean will be greater than 11.0 seconds can be calculated thus :

P(Z > 0.601) = 1 - P(Z < 0.601)

Using a normal distribution table ;

P(Z < 0.601) = 0.72608

P(Z > 0.601) = 1 - 0.72608 = 0.2739

Therefore, the approximate probability of the sample mean being greater than 11.0 seconds is 0.2739.

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24

step-by-step explanation:

the graph shows 30% of students walk. 80 students were questioned, so the number that said they walk is

30% × 80 = 0.30 × 80 = 24