Question

The mean caloric intake of males aged 20-39 years old is 2716 calories, with a standard deviation of 72.8 calories. Suppose a nutritionist analyzes a simple random sample of 35 males aged 20-39 and finds that their mean intake was 2750 calories. What is the probability that a random sample of 35 males aged 20-39 would result in a sample mean of 2750 calories or higher? Show All Work Pls.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 2716

standard deviation = $\sigma$ = 72.8

n = 35

$\mu$$\bar x$ = 2716

$\sigma$$\bar x$ = $\sigma$ / $\sqrt$ n = 72.8 / $\sqrt$35 = 12.305

P($\bar x$ >2750 ) = 1 - P($\bar x$ < 2750)

= 1 - P[($\bar x$ - $\mu$ $\bar x$ ) / $\sigma$ $\bar x$ < (2750-2716) /12.305 ]

= 1 - P(z <2.76 )

Using z table

= 1 - 0.9971

= 0.0029

probability= 0.0029