Question

Suppose that past records indicate that the probability that a new car will need a warranty repair in the first 90 days of use is 0.04. If a random sample of 400 new cars is selected, what is the probability that the proportion of new cars needing a warranty repair in the first 90 days will be: a. between 0.05 and 0.06? i.e. P(0.05 sp < 0.06) = ranswer to 4 decimal places). b. above 0.07? i.e. Plp > 0.07) = (round your answer to 5 decimal places) c. below 0.03? i.e. Plo < 0.03) = (round your answer to 4 decimal places)

Answer 1

Solution:

Given ,

p = 0.04 (population proportion)

1 - p = 1 - 0.04 = 0.96

n = 400 (sample size)

Let $\hat p$ be the sample proportion.

$\therefore$ The sampling distribution of $\hat p$ is approximately normal with

mean = $\mu_{\hat p}$ =  p = 0.04

SD = $\sigma_{\hat p}$ =     

p(1-P)/n

=  $\sqrt{0.04(1-0.04)/400}$

=  0.00979795897

a)

P(Sample proportion is between 0.05 and 0.06)

= P($\hat p$ < 0.06) - P($\hat p$ < 0.05)

=$P((\hat p-\mu_{\hat p})/\sigma_{\hat p} > (0.06-0.04)/ 0.00979795897) - P((\hat p-\mu_{\hat p})/\sigma_{\hat p} > (0.05-0.04)/ 0.00979795897) )$

= P(Z < 2.04) - P(Z < 1.02)

= 0.9793 - 0.8461

= 0.1332

b)

P(Sample proportion is above 0.07)

= P($\hat p$ > 0.07)

= $P((\hat p-\mu_{\hat p})/\sigma_{\hat p} > (0.07 -0.04)/0.00979795897 )$

= P(Z > 3.06)

= P(Z < -3.06)

= 0.0011 ...use z table

Answer : 0.00110

c)

P(Sample proportion is less than 0.03)

= P($\hat p$ < 0.03)

= P((\hat p-\mu_{\hat p})/\sigma_{\hat p} <(0.0.03 -0.04)/0.00979795897)

= P(Z < -1.02)

= 0.1539