Homework help, want to check my method for probability and hypothesis test

I would be most obliged if someone would look at my solution and advise if my method and answers are correct.

Mean lifetime of brand X bulbs is known to be 4,500 hours with standard deviation of 450 hours.

Part A

A light bulb is selected at random find the probability that it will last for at least 4,000 hours

Solution Part A:

At least 4,000 means 4,000 or more.

Zscore = (4000-4500)/450 = -1.111

want P(z >= -1.111) = P(z <= 1.11) = 0.8665 is this Answer corect or at least the correct method?

Part B

20 brand X bulbs are selected at random find probability that at least 18 of them will last for more than 4,000 hours.

Solution Part B:

Use binomial probability with probability of success for each bulb 0.8665 from part A above.

prob at least 18 = prob 18 + prob 19 + prob 20

using (n choose r) * p^r * q^(n-r) with n = 20, r = 18, 19 and 20, p = 0.8665 from above, q = 1-0.8665 = 0.1335. summing them up for r=18, 19, 20 gives Answer of 0.4891. Is this Answer correct or at least the correct method?

Part C

Manufacturer of brand Y bulbs claims his bulbs have the same mean of brand X = 4,5000. Considering brand X mean 4500, standard deviation of 450 is normal population. Sample of size 25 of brand Y is randomly pulled with a found mean of 4310 hours

Test at the 5% level of significance, the hypothesis that the mean lifespan of Brand Y is the same as the mean of X. State clearly what the manfacturere of brand X can conclude about brand Y.


Two tailed test, 2.5% in each tail. Student T distribution is not on course, only Z. The critical Z for upper tail i +1.96, for lower tail -1.96

H0 mean X = mean Y
Halternative mean Y does not equal Mean y

Appeal to the Central Limit Theorem, one sample Z test.

z= (4310-4500)/(450/suareroot(25)) = (-190)/(90) = -2.11

This is outside the interval -1.96...1.96, Reject the null hypotheses, Manufacturer X conclude that the means of brand y are different to the means of brand X. Is this Answer correct or at least the correct method?

Part D

Find the p-value of the test performed in part (C) above and explain what this value represents in the context of the question.


The area in the upper tail with z = 2.11 is from the tables (1 - 0.9826) = 0.0174. by symmetry the area in the lower tail is also 0.0174. The area in the tails is twice this = 0.0348. This is the p value. This tells us that the probability of obtaining the sample mean of 4,310 hours is 0.0348 if the brand Y mean is the same as brand X. The required level of significance was 5%, the obtained p value, 0.0348 = 3.48% is less than 5% so reject the null hypothesis. Is this Answer correct or at least the correct method?

Many thanks for taking the time to read this.