Question

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minutes per day that people spend standing or walking. Among mildly obese people, the mean number of minutes of daily activity (standing or walking) is approximately Normally distributed with 375 minutes and standard deviation 67 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with 523 minutes and standard deviation 107 minutes. A researcher records the minutes of activity for an SRS of 6 mildly obese people and an SRS of 6 lean people. (a) What is the probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes? (Enter your answer rounded to four decimal places.) probability: (b) What is the probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes? (Enter your answer rounded to four decimal places.) probability:

Answer 1

Answer:

(a) The probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes is 0.0495 .

(b) The probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes is 0.9909 .

Step-by-step explanation:

We are given that among mildly obese people, the mean number of minutes of daily activity (standing or walking) is approximately Normally distributed with Mean, $\mu$ = 375 minutes and standard deviation, $\sigma$ = 67 minutes.

The sample size for mildly obese people is n = 6.

The Z score normal probability for sample mean is given by;

    Z = $\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

(a) Let $Xbar$ = the mean number of minutes of daily activity of the 6 mildly obese people

So, Probability($Xbar$ > 420) = P($\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{420 - 375}{\frac{67}{\sqrt{6} } }$ )

                                              = P(Z > 1.65) = 1 - P(Z <= 1.65)

                                              = 1 - 0.95053 = 0.0495

Therefore, probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes is 0.0495.

(b) We are also given that the mean number of minutes of daily activity for lean people is approximately Normally distributed with Mean, $\mu$ = 523 minutes and standard deviation, $\sigma$ = 107 minutes.

The sample size for mildly lean people is n = 6.

Let $Xbar$ = the mean number of minutes of daily activity of the 6 mildly lean people

So, Probability($Xbar$ > 420) = P($\frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{420 - 523}{\frac{107}{\sqrt{6} } }$ )

                                              = P(Z > -2.36) = P(Z < 2.36) = 0.9909

Therefore, probability that the mean number of minutes of daily activity of the 6 mildly lean people exceeds 420 minutes is 0.9909.

We have used Z table for calculating above probabilities.

                                             

Answer 2

Answer:

(a) The mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes is 0.0495.

(b) The mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes is 0.9909.

Step-by-step explanation:

Let X = amount of time mildly obese people do daily activity and Y = amount of time lean people do daily activity.

The random variable $X\sim N(375, 67^{2})$ and $Y\sim N(523, 107^{2})$.

The sample size of mildly obese and lean people selected is, n = 6.

(a)

Compute the probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes as follows:

$P(\bar X->420)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{420-375}{67/\sqrt{6}} )\\=P(Z>1.65)\\=1-P(Z$

*Use the z-table for the probability.

Thus, the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes is 0.0495.

(b)

Compute the probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes as follows:

$P(\bar Y->420)=P(\frac{\bar Y-\mu}{\sigma/\sqrt{n}}>\frac{420-523}{107/\sqrt{6}} )\\=P(Z>-2.36)\\=P(Z$

*Use the z-table for the probability.

Thus, the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes is 0.9909.