Question

A happiness survey with normally distributed scores with mean 5.5 and standard deviation 2.3 was administered to IT workers in healthcare. a) Find the probability that a randomly selected participant's response was greater than 5. 2- 5-5.9 2= -0.2174 0.4129 2.3 -0.4129 -0.5871 b) Find the probability that a randomly selected participant's response was between 4.5 and 6.5. 6.5-5.5 2-0.434783 22.30 = -0.434783 0.6604-0-5536 = 2.3 0.6664 0.3336 [0.3328 c) Find the probability that the mean of a sample of 16 selected participant's response was between 4.5 and 6.5.

Answer 1

a) 58.71% probability that a randomly selected participant's response was greater than 5

b) 33.28% probability that a randomly selected participant's response was between 4.5 and 6.5.

c) 91.82% probability that the mean of a sample of 16 selected participant's response was between 4.5 and 6.5.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, a large sample size can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 5.5, \sigma = 2.3`

a) Find the probability that a randomly selected participant's response was greater than 5

This probability is 1 subtracted by the pvalue of Z when X = 5. So

`Z = (X - \mu)/(\sigma)`

`Z = (5 - 5.5)/(2.3)`

`Z = -0.22`

`Z = -0.22` has a pvalue of 0.4129.

1 - 0.4129 = 0.5871

58.71% probability that a randomly selected participant's response was greater than 5

b) Find the probability that a randomly selected participant's response was between 4.5 and 6.5.

This probability is the pvalue of Z when X = 6.5 subtracted by the pvalue of Z when X = 4.5. So

X = 6.5

`Z = (X - \mu)/(\sigma)`

`Z = (6.5 - 5.5)/(2.3)`

`Z = 0.43`

`Z = 0.43` has a pvalue of 0.6664.

X = 4.5

`Z = (X - \mu)/(\sigma)`

`Z = (4.5 - 5.5)/(2.3)`

`Z = -0.43`

`Z = -0.43` has a pvalue of 0.3336

0.6664 - 0.3336 = 0.3328

33.28% probability that a randomly selected participant's response was between 4.5 and 6.5.

c) Find the probability that the mean of a sample of 16 selected participant's response was between 4.5 and 6.5.

Now we have
`n = 16, s = (2.3)/(√(16)) = 0.575`

This probability is the pvalue of Z when X = 6.5 subtracted by the pvalue of Z when X = 4.5. So

X = 6.5

`Z = (X - \mu)/(\sigma)`

Due to the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (6.5 - 5.5)/(0.575)`

`Z = 1.74`

`Z = 1.74` has a pvalue of 0.9591.

X = 4.5

`Z = (X - \mu)/(s)`

`Z = (4.5 - 5.5)/(0.575)`

`Z = -1.74`

`Z = -1.74` has a pvalue of 0.0409

0.9591 - 0.0409 = 0.9182

91.82% probability that the mean of a sample of 16 selected participant's response was between 4.5 and 6.5.