Question

The reading speed of second grade students is approximately normal, with a mean of 90 words per minute (wpm) and a standard deviation of 10 wpm. Complete parts (a) through (f). (a) What is the probability a randomly selected student will read more than 95 words per minute? The probability is (Round to four decimal places as needed.) (b) What is the probability that a random sample of 11 second grade students results in a mean reading rate of more than 95 words per minute? The probability is (Round to four decimal places as needed.) (c) What is the probability that a random sample of 22 second grade students results in a mean reading rate of more than 95 words per minute? The probability is (Round to four decimal places as needed.) (d) What effect does increasing the sample size have on the probability? Provide an explanation for this result. O A. Increasing the sample size increases the probability because of decreases as n increases. OB. Increasing the sample size decreases the probability because o increases as n increases. OC. Increasing the sample size increases the probability because of increases as n increases. OD. Increasing the sample size decreases the probability because o: decreases as n increases.

Answer 1

Solution :

Given that ,

mean = $\mu$ = 90

standard deviation = $\sigma$ = 10

P(x >95 ) = 1 - P(x <95 )

= 1 - P[(x - $\mu$ ) / $\sigma$ < (95 - 90) /10 ]

= 1 - P(z < 0.5)

Using z table,

= 1 -0.6915

=0.3085

(B)

n = 11

$\mu$_{$\bar x$} = 90

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 10/ $\sqrt$ 11 = 3.0151

P($\bar x$ > 95) = 1 - P($\bar x$ <95 )

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (95 - 90) / 3.0151]

= 1 - P(z <1.66 )

Using z table,    

= 1 - 0.9515

= 0.0485

(C)

n = 22

$\mu$_{$\bar x$} = 90

$\sigma$_{$\bar x$} = $\sigma$ / $\sqrt$ n = 10/ $\sqrt$ 22 = 2.1320

P($\bar x$ > 95) = 1 - P($\bar x$ <95 )

= 1 - P[($\bar x$ - $\mu$ _{$\bar x$} ) / $\sigma$ _{$\bar x$} < (95 - 90) / 2.1320]

= 1 - P(z <2.35)

Using z table,    

= 1 - 0.9906

=0.0094