Question

a company has developed an \entry - shot\ mower that cuts the mowing width of a lawnmower. the company claims that the probability that the mower will start on any push of the button is 0.9. assume for now that this claim is true. one user made 10 uses of the mower, let y = the number of times it starts on the first push of the button.

(a) find the probability of at most 6 starts in 10 attempts if the company’s claim is true.

(rounded to 4 decimal places.
leave your answer as directed here.)

(b) suppose that the mower only starts on 6 of the 10 attempts. does this give convincing evidence that the company’s claim is exaggerated? explain your reasoning.

it is that the mower would start 6 times or fewer out of 10 attempts by chance alone, so there convincing evidence that the company’s claim is exaggerated.

(c) calculate the expected value of y.

(do not round.)

interpret the expected value of y.

if 10 mower uses are selected, we expect that the mower will start about of them times, on average.

(d) calculate the standard deviation of y.

(rounded to 2 decimal places.)

interpret the standard deviation of y.

if 10 mower uses are selected, the number of times that the mower will start would typically vary from the mean of by about starts.

Explanation:

Step1: Identify the distribution

This is a binomial distribution problem where \( n = 10 \) (number of attempts), \( p = 0.9 \) (probability of success), and \( Y \) is the number of successes. The binomial probability formula is \( P(Y = k)=\binom{n}{k}p^{k}(1 - p)^{n - k} \), and for "at most 6", we calculate \( P(Y\leq6)=\sum_{k = 0}^{6}\binom{10}{k}(0.9)^{k}(0.1)^{10 - k} \).

Step2: Calculate the sum

We can use a calculator or software to compute the sum. Alternatively, we know that \( P(Y\leq6)=1 - P(Y\geq7) \). Calculating \( P(Y = 7)=\binom{10}{7}(0.9)^{7}(0.1)^{3}=\frac{10!}{7!3!}(0.4782969)(0.001)=120\times0.4782969\times0.001 = 0.0573956 \)
\( P(Y = 8)=\binom{10}{8}(0.9)^{8}(0.1)^{2}=\frac{10!}{8!2!}(0.43046721)(0.01)=45\times0.43046721\times0.01 = 0.19371024 \)
\( P(Y = 9)=\binom{10}{9}(0.9)^{9}(0.1)^{1}=10\times(0.387420489)\times0.1 = 0.387420489 \)
\( P(Y = 10)=\binom{10}{10}(0.9)^{10}(0.1)^{0}=1\times(0.3486784401)\times1 = 0.3486784401 \)
Summing these: \( P(Y\geq7)=0.0573956 + 0.19371024+0.387420489 + 0.3486784401=0.9872047691 \)
Then \( P(Y\leq6)=1 - 0.9872047691 = 0.0127952309\approx0.0128 \)

We found in part (a) that \( P(Y\leq6)\approx0.0128 \), which is a small probability (less than 0.05, typically). A small probability event (starting 6 or fewer times) has occurred, so it is unlikely that the company's claim ( \( p = 0.9 \)) is true. So it is unlikely that the mower would start 6 times or fewer out of 10 attempts by chance alone, so there is convincing evidence that the company’s claim is exaggerated.

Step1: Recall the expected value formula for binomial distribution

The expected value of a binomial random variable \( Y \) with parameters \( n \) and \( p \) is \( E(Y)=np \).

Step2: Substitute the values

Here, \( n = 10 \) and \( p = 0.9 \), so \( E(Y)=10\times0.9 = 9 \)

Step1: Recall the standard deviation formula for binomial distribution

The standard deviation of a binomial random variable \( Y \) is \( \sigma=\sqrt{np(1 - p)} \)

Step2: Substitute the values

\( n = 10 \), \( p = 0.9 \), \( 1 - p = 0.1 \). So \( \sigma=\sqrt{10\times0.9\times0.1}=\sqrt{0.9}\approx0.94868\approx0.95 \)

Answer:

\( 0.0128 \)

Part (b)