Acompany makes auto batteries. they claim that 84 % of their ll70 batteries are good for 70 months or longer. assume that this claim is true. let p overscript ^ endscripts be the proportion in a random sample of 60 such batteries that are good for 70 months or more. a. what is the probability that this sample proportion is within 0.03 of the population proportion? round your answer to two decimal places. b.what is the probability that this sample proportion is less than the population proportion by 0.05 or more? round your answer to two decimal places.

a) 0.47

b) 0.14

Step-by-step explanation:

Since the company claim that 84% of the LL70 batteries are good 70 months or longer, let the population proportion(p) = 84%

p= 0.84

q = 1 - p

q= 1 - 0.84

q= 0.16

n = 60

Variance of population = pq/n

= (0.84*0.16) / 60

= 0.1344/60

= 0.00224

Standard deviation = √variance

= √0.00224

= 0.0473

Let P be the proportion of a sample of 60

a) Probability that the sample is within 0.03 is Pr(0.81 < P < 0.87)

= [(0.81 - 0.84)/0.0473 < Z < (0.87 - 0.84)/0.0473

= [-0.03/0.0473 < Z < 0.03/0.0473]

= P(-0.6342 < Z < 0.6342)

= P(-0.6342 < Z < 0.6342) = 2P( 0 < Z < 0.6342)

From the normal distribution table

0.6342 = 0.2370

Φ(Z) = 0.2370

P(-0.6342 < Z < 0.6342) = 2(0.2370)

= 0.4740

= 0.47(approximate to 2 d. p)

b) P(p-0.84<0.05)

= P( Z< -0.05/0.0473)

= P(Z < -1.057)

From the table, 1.057 = 0.3574

Φ(Z) = 0.3574

Recall that if Z is negative,

Pr(X<a) = 0.5 - Φ(Z)

P(Z < -1.057) = 0.5 - 0.3574

= 0.1426

= 0.14(approximate to 2 d.p)

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